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  • GRUB is not Booting Correctly

    - by msknapp
    I have a PC with three hard disks. Windows 7 is installed on the first, Ubuntu 14.04 is installed on the third. After I re-booted, it went straight to Windows 7. So I tried explicitly telling my PC to boot using the third hard disk, but that just takes me to the grub rescue prompt. I followed Scott Severence's instructions here to try and recover. Essentially, I updated grub, reinstalled grub, and then updated it again. After re-booting, absolutely nothing had changed. So instead I tried using the boot-repair tool. In the past it had failed for me, saying that I had programs running and it could not unmount drives, when I was running nothing. I never figured out how to solve that problem, but it went away when I bought another hard drive and used that for my Ubuntu installation, I don't know why. In any case, I ran the boot-repair tool and this time it said it was successful. First time for everything right? I re-booted, only to be taken straight to the grub rescue prompt. So I changed my BIOS settings to use the third hard disk for boot start up. That is the same hard drive where I have Ubuntu and grub installed, and the same one that the grub-repair tool told me to use. It still took me straight to the grub rescue prompt. So I went from not being able to boot Ubuntu, to not being able to boot either OS installed on my system. Thanks boot-repair! Boot repair gave me this URL for future troubleshooting: http://paste.ubuntu.com/8131669 When I try to boot from the third hard disk, this is my console: Loading Operating System ... error: attempt to read or write outside of disk 'hd0'. Entering rescue mode... grub rescue> grub rescue> set cmdpath=(hd0) prefix=(hd0,gpt2)/boot/grub root=hd0,gpt2 grub rescue> ls (hd0) (hd0,gpt3) (hd0,gpt2) (hd0,gpt1) (hd1) (hd2) (hd2,gpt2) (hd2,gpt1) (hd3) Those values look correct to me. I have also experimented with changing some of those values, but 'insmod normal' always throws the same error. Somebody please tell me how to fix this. I have tried everything, reinstalling grub, and running boot-repair. =========================== Update: I think the problem might be that the ubuntu installer did not partition my hard disk correctly. I booted from live USB and then launched gparted and looked at how it partitioned things. This is what gparted says: Partition, File System, Size, Used, Unused, Flags /dev/sda1 (!), unknown, 1.00 MiB, ---, ---, bios_grub /dev/sda2, ext4, 2.71 TiB, 47.30 GiB, 2.67 TiB, /dev/sda3, linux-swap, 16.00 GiB, 0.00 B, 16.00 GiB, So that first line looks problematic. It is supposed to be the /boot partition. However, it was given only 1 MiB? I am assuming that MiB is actually supposed to mean megabyte, no idea why that 'i' is there. It also says the file system is unknown. I read the answer by andrew here, and he says he had to do a custom install, explicitly configuring the boot partition. So I think that maybe Ubuntu's installer has a bug in it, where it does not set up the boot partition correctly if you are not installing on the first hard disk in your computer. I am going to try reinstalling with a custom partition scheme. I read elsewhere (askubuntu won't let me post another link) that I don't even need a /boot partition any more. So instead of following Andrew's instructions ver batim, I'm first going to try having just two partitions: one for /, and another for my 16GB swap space. Both as primary partitions. The first will be formatted as ext4. If that doesn't work, I may try again using /boot. ======================== So I did my custom install with no /boot partition, and it did not work. When I rebooted, I had an error message saying that some address did not exist. So for the hundredth time, I booted from the live USB, and ran boot-repair. Now I get this message GPT detected. Please create a BIOS-Boot partition (>1MB, unformatted filesystem, bios_grub flag). This can be performed via tools such as Gparted. Then try again. I feel like I'm running in circles and nobody will help me.

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  • SQL Quey slow in .NET application but instantaneous in SQL Server Management Studio

    - by user203882
    Here is the SQL SELECT tal.TrustAccountValue FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = 70402 AND ta.TrustAccountID = 117249 AND tal.trustaccountlogid = ( SELECT MAX (tal.trustaccountlogid) FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = 70402 AND ta.TrustAccountID = 117249 AND tal.TrustAccountLogDate < '3/1/2010 12:00:00 AM' ) Basicaly there is a Users table a TrustAccount table and a TrustAccountLog table. Users: Contains users and their details TrustAccount: A User can have multiple TrustAccounts. TrustAccountLog: Contains an audit of all TrustAccount "movements". A TrustAccount is associated with multiple TrustAccountLog entries. Now this query executes in milliseconds inside SQL Server Management Studio, but for some strange reason it takes forever in my C# app and even timesout (120s) sometimes. Here is the code in a nutshell. It gets called multiple times in a loop and the statement gets prepared. cmd.CommandTimeout = Configuration.DBTimeout; cmd.CommandText = "SELECT tal.TrustAccountValue FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = @UserID1 AND ta.TrustAccountID = @TrustAccountID1 AND tal.trustaccountlogid = (SELECT MAX (tal.trustaccountlogid) FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = @UserID2 AND ta.TrustAccountID = @TrustAccountID2 AND tal.TrustAccountLogDate < @TrustAccountLogDate2 ))"; cmd.Parameters.Add("@TrustAccountID1", SqlDbType.Int).Value = trustAccountId; cmd.Parameters.Add("@UserID1", SqlDbType.Int).Value = userId; cmd.Parameters.Add("@TrustAccountID2", SqlDbType.Int).Value = trustAccountId; cmd.Parameters.Add("@UserID2", SqlDbType.Int).Value = userId; cmd.Parameters.Add("@TrustAccountLogDate2", SqlDbType.DateTime).Value =TrustAccountLogDate; // And then... reader = cmd.ExecuteReader(); if (reader.Read()) { double value = (double)reader.GetValue(0); if (System.Double.IsNaN(value)) return 0; else return value; } else return 0;

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  • SQL Query slow in .NET application but instantaneous in SQL Server Management Studio

    - by user203882
    Here is the SQL SELECT tal.TrustAccountValue FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = 70402 AND ta.TrustAccountID = 117249 AND tal.trustaccountlogid = ( SELECT MAX (tal.trustaccountlogid) FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = 70402 AND ta.TrustAccountID = 117249 AND tal.TrustAccountLogDate < '3/1/2010 12:00:00 AM' ) Basicaly there is a Users table a TrustAccount table and a TrustAccountLog table. Users: Contains users and their details TrustAccount: A User can have multiple TrustAccounts. TrustAccountLog: Contains an audit of all TrustAccount "movements". A TrustAccount is associated with multiple TrustAccountLog entries. Now this query executes in milliseconds inside SQL Server Management Studio, but for some strange reason it takes forever in my C# app and even timesout (120s) sometimes. Here is the code in a nutshell. It gets called multiple times in a loop and the statement gets prepared. cmd.CommandTimeout = Configuration.DBTimeout; cmd.CommandText = "SELECT tal.TrustAccountValue FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = @UserID1 AND ta.TrustAccountID = @TrustAccountID1 AND tal.trustaccountlogid = (SELECT MAX (tal.trustaccountlogid) FROM TrustAccountLog AS tal INNER JOIN TrustAccount ta ON ta.TrustAccountID = tal.TrustAccountID INNER JOIN Users usr ON usr.UserID = ta.UserID WHERE usr.UserID = @UserID2 AND ta.TrustAccountID = @TrustAccountID2 AND tal.TrustAccountLogDate < @TrustAccountLogDate2 ))"; cmd.Parameters.Add("@TrustAccountID1", SqlDbType.Int).Value = trustAccountId; cmd.Parameters.Add("@UserID1", SqlDbType.Int).Value = userId; cmd.Parameters.Add("@TrustAccountID2", SqlDbType.Int).Value = trustAccountId; cmd.Parameters.Add("@UserID2", SqlDbType.Int).Value = userId; cmd.Parameters.Add("@TrustAccountLogDate2", SqlDbType.DateTime).Value =TrustAccountLogDate; // And then... reader = cmd.ExecuteReader(); if (reader.Read()) { double value = (double)reader.GetValue(0); if (System.Double.IsNaN(value)) return 0; else return value; } else return 0;

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  • How can I choose different hints for different joins for a single table in a query hint?

    - by RenderIn
    Suppose I have the following query: select * from A, B, C, D where A.x = B.x and B.y = C.y and A.z = D.z I have indexes on A.x and B.x and B.y and C.y and D.z There is no index on A.z. How can I give a hint to this query to use an INDEX hint on A.x but a USE_HASH hint on A.z? It seems like hints only take the table name, not the specific join, so when using a single table with multiple joins I can only specify a single strategy for all of them. Alternative, suppose I'm using a LEADING or ORDERED hint on the above query. Both of these hints only take a table name as well, so how can I ensure that the A.x = B.x join takes place before the A.z = D.z one? I realize in this case I could list D first, but imagine D subsequently joins to E and that the D-E join is the last one I want in the entire query. A third configuration -- Suppose I want the A.x join to be the first of the entire query, and I want the A.z join to be the last one. How can I use a hint to have a single join from A to take place, followed by the B-C join, and the A-D join last?

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  • Mysql partitioning: Partitions outside of date range is included

    - by Sturlum
    Hi, I have just tried to configure partitions based on date, but it seems that mysql still includes a partition with no relevant data. It will use the relevant partition but also include the oldest for some reason. Am I doing it wrong? The version is 5.1.44 (MyISAM) I first added a few partitions based on "day", which is of type "date" ALTER TABLE ptest PARTITION BY RANGE(TO_DAYS(day)) ( PARTITION p1 VALUES LESS THAN (TO_DAYS('2009-08-01')), PARTITION p2 VALUES LESS THAN (TO_DAYS('2009-11-01')), PARTITION p3 VALUES LESS THAN (TO_DAYS('2010-02-01')), PARTITION p4 VALUES LESS THAN (TO_DAYS('2010-05-01')) ); After a query, I find that it uses the "old" partition, that should not contain any relevant data. mysql> explain partitions select * from ptest where day between '2010-03-11' and '2010-03-12'; +----+-------------+------------+------------+-------+---------------+------+---------+------+------+-------------+ | id | select_type | table | partitions | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+------------+------------+-------+---------------+------+---------+------+------+-------------+ | 1 | SIMPLE | ptest | p1,p4 | range | day | day | 3 | NULL | 79 | Using where | +----+-------------+------------+------------+-------+---------------+------+---------+------+------+-------------+ When I select a single day, it works: mysql> explain partitions select * from ptest where day = '2010-03-11'; +----+-------------+------------+------------+------+---------------+------+---------+-------+------+-------+ | id | select_type | table | partitions | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+------------+------------+------+---------------+------+---------+-------+------+-------+ | 1 | SIMPLE | ptest | p4 | ref | day | day | 3 | const | 39 | | +----+-------------+------------+------------+------+---------------+------+---------+-------+------+-------+

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  • How do I write this GROUP BY in mysql UNION query

    - by user1652368
    Trying to group the results of two queries together. When I run this query: SELECT pr_id, pr_sbtcode, pr_sdesc, od_quantity, od_amount FROM ( SELECT `bgProducts`.`pr_id`, `bgProducts`.`pr_sbtcode`, `bgProducts`.`pr_sdesc`, SUM(`od_quantity`) AS `od_quantity`, SUM(`od_amount`) AS `od_amount`, MIN(UNIX_TIMESTAMP(`or_date`)) AS `or_date` FROM `bgOrderMain` JOIN `bgOrderData` JOIN `bgProducts` WHERE `bgOrderMain`.`or_id` = `bgOrderData`.`or_id` AND `od_pr` = `pr_id` AND UNIX_TIMESTAMP(`or_date`) >= '1262322000' AND UNIX_TIMESTAMP(`or_date`) <= '1346990399' AND (`pr_id` = '415' OR `pr_id` = '1088') GROUP BY `bgProducts`.`pr_id` UNION SELECT `bgProducts`.`pr_id`, `bgProducts`.`pr_sbtcode`, `bgProducts`.`pr_sdesc`,SUM(`od_quantity`) AS `od_quantity`, SUM(`od_amount`) AS `od_amount`, MIN(UNIX_TIMESTAMP(`or_date`)) AS `or_date` FROM `npOrderMain` JOIN `npOrderData` JOIN `bgProducts` WHERE `npOrderMain`.`or_id` = `npOrderData`.`or_id` AND `od_pr` = `pr_id` AND UNIX_TIMESTAMP(`or_date`) >= '1262322000' AND UNIX_TIMESTAMP(`or_date`) <= '1346990399' AND (`pr_id` = '415' OR `pr_id` = '1088') GROUP BY `bgProducts`.`pr_id` ) TEMPTABLE3; it produces this result +-------+------------+--------------------------+-------------+-----------+ | pr_id | pr_sbtcode | pr_sdesc | od_quantity | od_amount +-------+------------+--------------------------+-------------+-----------+ | 415 | NP13 | Product 13 | 5 | 125 | 1088 | NPAW | Product AW | 4 | 100 | 415 | NP13 | Product 13 | 5 | 125 | 1088 | NPAW | Product AW | 2 | 50 +-------+------------+--------------------------+-------------+-----------+</pre> What I want to get a result that combines those into 2 lines: +-------+------------+--------------------------+-------------+-----------+ | pr_id | pr_sbtcode | pr_sdesc | od_quantity | od_amount +-------+------------+--------------------------+-------------+-----------+ | 415 | NP13 | Product 13 | 10 | 250 | 1088 | NPAW | Product AW | 6 | 150 +-------+------------+--------------------------+-------------+-----------+</pre> So I added GROUP BY pr_id to the end of the query: SELECT pr_id, pr_sbtcode, pr_sdesc, od_quantity, od_amount FROM ( SELECT `bgProducts`.`pr_id`, `bgProducts`.`pr_sbtcode`, `bgProducts`.`pr_sdesc`, SUM(`od_quantity`) AS `od_quantity`, SUM(`od_amount`) AS `od_amount`, MIN(UNIX_TIMESTAMP(`or_date`)) AS `or_date` FROM `bgOrderMain` JOIN `bgOrderData` JOIN `bgProducts` WHERE `bgOrderMain`.`or_id` = `bgOrderData`.`or_id` AND `od_pr` = `pr_id` AND UNIX_TIMESTAMP(`or_date`) >= '1262322000' AND UNIX_TIMESTAMP(`or_date`) <= '1346990399' AND (`pr_id` = '415' OR `pr_id` = '1088') GROUP BY `bgProducts`.`pr_id` UNION SELECT `bgProducts`.`pr_id`, `bgProducts`.`pr_sbtcode`, `bgProducts`.`pr_sdesc`,SUM(`od_quantity`) AS `od_quantity`, SUM(`od_amount`) AS `od_amount`, MIN(UNIX_TIMESTAMP(`or_date`)) AS `or_date` FROM `npOrderMain` JOIN `npOrderData` JOIN `bgProducts` WHERE `npOrderMain`.`or_id` = `npOrderData`.`or_id` AND `od_pr` = `pr_id` AND UNIX_TIMESTAMP(`or_date`) >= '1262322000' AND UNIX_TIMESTAMP(`or_date`) <= '1346990399' AND (`pr_id` = '415' OR `pr_id` = '1088') GROUP BY `bgProducts`.`pr_id` ) TEMPTABLE3 GROUP BY pr_id; But that just gives me this: +-------+------------+--------------------------+-------------+-----------+ | pr_id | pr_sbtcode | pr_sdesc | od_quantity | od_amount +-------+------------+--------------------------+-------------+-----------+ | 415 | NP13 | Product 13 | 5 | 125 | 1088 | NPAW | Product AW | 4 | 100 +-------+------------+--------------------------+-------------+-----------+ What am I missing here??

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  • Linq to Entities Joins

    - by Bob Avallone
    I have a question about joins when using Linq to Entities. According to the documentation the use on the join without a qualifier performs like a left outer join. However when I execute the code below, I get a count returned of zero. But if I comment out the three join lines I get a count of 1. That would indicate that the join are acting as inner join. I have two questions. One which is right inner or outer as the default? Second how do I do the other one i.e. inner or outer? The key words on inner and outer do not work. var nprs = (from n in db.FMCSA_NPR join u in db.FMCSA_USER on n.CREATED_BY equals u.ID join t in db.LKUP_NPR_TYPE on n.NPR_TYPE_ID equals t.ID join s in db.LKUP_AUDIT_STATUS on n.NPR_STATUS_ID equals s.ID where n.ROLE_ID == pRoleId && n.OWNER_ID == pOwnerId && n.NPR_STATUS_ID == pNPRStatusId && n.ACTIVE == pActive select n).ToList(); if (nprs.Count() == 0) return null;

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  • Translate an IQueryable instance to LINQ syntax in a string

    - by James Dunne
    I would like to find out if anyone has existing work surrounding formatting an IQueryable instance back into a LINQ C# syntax inside a string. It'd be a nice-to-have feature for an internal LINQ-to-SQL auditing framework I'm building. Once my framework gets the IQueryable instance from a data repository method, I'd like to output something like: This LINQ query: from ce in db.EiClassEnrollment join c in db.EiCourse on ce.CourseID equals c.CourseID join cl in db.EiClass on ce.ClassID equals cl.ClassID join t in db.EiTerm on ce.TermID equals t.TermID join st in db.EiStaff on cl.Instructor equals st.StaffID where (ce.StudentID == studentID) && (ce.TermID == termID) && (cl.Campus == campusID) select new { ce, cl, t, c, st }; Generates the following LINQ-to-SQL query: DECLARE @p0 int; DECLARE @p1 int; DECLARE @p2 int; SET @p0 = 777; SET @p1 = 778; SET @p2 = 779; SELECT [t0].[ClassEnrollmentID], ..., [t4].[Name] FROM [dbo].[ei_ClassEnrollment] AS [t0] INNER JOIN [dbo].[ei_Course] AS [t1] ON [t0].[CourseID] = [t1].[CourseID] INNER JOIN [dbo].[ei_Class] AS [t2] ON [t0].[ClassID] = [t2].[ClassID] INNER JOIN [dbo].[ei_Term] AS [t3] ON [t0].[TermID] = [t3].[TermID] INNER JOIN [dbo].[ei_Staff] AS [t4] ON [t2].[Instructor] = [t4].[StaffID] WHERE ([t0].[StudentID] = @p0) AND ([t0].[TermID] = @p1) AND ([t2].[Campus] = @p2) I already have the SQL output working as you can see. I just need to find a way to get the IQueryable to translate into a string representing its original LINQ syntax (with an acceptable translation loss). I'm not afraid of writing it myself, but I'd like to see if anyone else has done this first.

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  • Reorganize Primary/Recovery Partitions and then install Ubuntu Netbook Remix?

    - by Wesley
    Hi all, I have a Samsung N120 netbook (with upgraded 2GB RAM). I recently got a new hard drive and motherboard for it because the original parts were faulty. However, when I got it back, whoever was working on it decided to make my primary partition ~40GB in size, compared to (what appears to be) the recovery partition, which is around 100GB in size. Firstly, I want to make the recovery partition much smaller (around ~10GB or smaller, if possible) and then make my primary partition fill the rest of the space. After that, I want to install Ubuntu Netbook Remix 9.10. I don't know if this is installed within Windows like Ubuntu 9.10 but I want both XP and UNR on my machine at the same time. So basically... how do I resize my primary and recovery partitions such that the recovery is about 10GB or less and the primary fills the rest? Secondly, is UNR installed within Windows? (If not, would it create its own partition on my hard drive, along with the XP partition, and the recovery partition?) Thanks in advance.

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  • Cannot Install Bootcamp on Mac Fusion Drive

    - by user3377019
    I have two drive installed in my mid-2013 Macbook pro. It is running osx maverick. I set the two drives (an ssd and a regular HD) up using the fusion drive (following the diy fusion drive guides out there). I went ahead and created a 40gb partition to host my windows install. I removed the SSD, installed windows, then reinstalled the SSD. As soon as the ssd was placed back in the bootcamp partition stopped booting. I get a blinking cursor on a black screen. I checked out the partition info in disk utility and it appears that the windows partition is not marked bootable. Below is some info I managed to gather. I am wondering if there is a way to fix the partition table so my bootcamp will boot. /dev/disk0 #: TYPE NAME SIZE IDENTIFIER 0: GUID_partition_scheme *120.0 GB disk0 1: EFI EFI 209.7 MB disk0s1 2: Apple_CoreStorage 119.7 GB disk0s2 3: Apple_Boot Boot OS X 134.2 MB disk0s3 /dev/disk1 #: TYPE NAME SIZE IDENTIFIER 0: GUID_partition_scheme *500.1 GB disk1 1: EFI EFI 209.7 MB disk1s1 2: Microsoft Basic Data BOOTCAMP 40.0 GB disk1s2 3: Apple_CoreStorage 459.2 GB disk1s3 4: Apple_Boot Boot OS X 650.0 MB disk1s4 /dev/disk2 #: TYPE NAME SIZE IDENTIFIER 0: Apple_HFS Macintosh HD *573.4 GB disk2 Name : BOOTCAMP Type : Partition Disk Identifier : disk1s2 Mount Point : /Volumes/BOOTCAMP File System : Windows NT File System (NTFS) Connection Bus : SATA Device Tree : IODeviceTree:/PCI0@0/SATA@1F,2/PRT1@1/PMP@0 Writable : No Universal Unique Identifier : 584BAED6-4C46-4F18-93B3-957F6E27003C Capacity : 40 GB (39,998,980,096 Bytes) Free Space : 16.34 GB (16,339,972,096 Bytes) Used : 23.66 GB (23,659,003,904 Bytes) Number of Files : 86,424 Number of Folders : 0 Owners Enabled : No Can Turn Owners Off : No Can Be Formatted : No Bootable : No Supports Journaling : No Journaled : No Disk Number : 1 Partition Number : 2

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  • Windows 7 boots to black screen with blinking cursor

    - by murgatroid99
    I have an Alienware M17x that dual boots into Ubuntu 11.04 and Windows 7 Home Premium. Currently, the computer starts at the GRUB loader and will boot into Ubuntu, but if I try to boot into Windows, I immediately get a black screen with a blinking cursor in the upper left corner. The output of fdisk -l is Device Boot Start End Blocks Id System /dev/dm-0p1 1 5 40131 de Dell Utility Partition 1 does not start on physical sector boundary. /dev/dm-0p2 6 1918 15360000 7 HPFS/NTFS Partition 2 does not start on physical sector boundary. /dev/dm-0p3 * 1918 64772 504878877+ 7 HPFS/NTFS Partition 3 does not start on physical sector boundary. /dev/dm-0p4 64772 77827 104858625 5 Extended Partition 4 does not start on physical sector boundary. /dev/dm-0p5 64772 67204 19531008 83 Linux /dev/dm-0p6 67204 74498 58593536 83 Linux /dev/dm-0p7 74498 77577 24731648 83 Linux /dev/dm-0p8 77578 77827 2000128 82 Linux swap / Solaris I have used the Windows rescue CD, and run the automatic error fixer until it finds no errors. I have run chkdsk /R on both the main Windows 7 (/dev/dm-0p3) partition and the recovery partition (/dev/dm-0p2). I set the main Windows 7 partition to be active. I also tried running in the recovery console the commands bootrec /fixmbr bootrec /fixboot bootrec /rebuildbcd None of these helped and the last set of commands deletes grub, which I then have to reinstall from Ubuntu. I think the last thing I did in windows before this started was install the newest ATI driver for my video card. This would suggest using system restore, and I actually had a restore point earlier (after the problem started), but after whatever I did that restore point does not appear in the list on the recovery disk any more, so I cannot do a system restore. Is there anything else I can try to make Windows boot properly again? Edit: Running the suggested commands bootsect /nt60 c: bcdboot c:\windows /s c: was also ineffective.

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  • Partitioning & Linux

    - by Zac
    Every tutorial on Linux-based partitioning schemes (or, just partitioning in general) will tell you that a PC can have either 4 primary partitions, or 3 primaries and 1 extended. They will all also tell you that Linux (in my case, Ubuntu) can be installed on either. It's also come to my attention that it is not too atypical for FHS directories, such as usr/, tmp/, etc/, home/ or var/ to be mounted separately on other partitions. Several questions I am unable to find the answers to, purely for my own edification: (1) By "PC", are we really talking about common PC disk types, like IDE or SATA? I guess I'm wondering why PC uses are limited to 4 primaries or 3 primaries + 1 extended (2) I'm choking on some basic OS concepts: it is said that a partition can be mounted by a file system or an OS. So I assume this means I can somehow instruct Ubuntu to mount to 1 partition, and then any part of, say, ReiserFS, to be mounted to another partition? How? (3)(a) What about creating swap partitions? Is there too much of a good thing with swap partitioning? If I have 4GB RAM over 320GB disk, what should my swap partition size be, and why? (3)(b) Are swap files the only way to create swap partitions? Wouldn't a Linux partitioning utility allow me to define a partition as being for virtual memory only? (4) Why are partitions limited to being "mounted" by just OSes and file systems? Why couldn't I write a program to take up its own, say, 512 MB partition, and then have it invoked or uses by an OS installed on another partition? Thanks for shedding any light here... not critical that I know this stuff, but it's got me thinking incessantly. And when I think incessantly, I...can't......sleep....

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  • Regarding partitions for dual-booting Ubuntu with pre-existing Windows 7

    - by Shasteriskt
    I have zero actual experience with configuring disk partitions and the stuff I have read for the past few hours have been confusing me a bit, so please bear with me. First of all, I'd like to explain what I'm setting to achieve: Windows 7 with: C:\ Windows 7 (pre-existing installation) D:\ Data (Already exists and has files already) Ubuntu 11 - Does not exist yet, but I already have a LiveCD in hand. \root directory for Ubuntu \home on its own partition I plan \swap on its own partition with around 8GB Here is the current situation: I have a single 500 GB hard-disk with Windows 7 x64 installed, and the current partition schemes is as follows: System Reserved: 100 MB (Primary, Active) C: 100 GB - Where Windows 7 is installed (Primary) D: 365 GB - Where my files are located, LOTS of free space (Primary) Now, I would like to shrink my D: drive and create around 40 GB of unallocated disk space for the Ubuntu installation, but here what's confusing me a bit: I'm thinking I would create an extended partition and subdivide it into 3 logical partitions for the Ubuntu setup I had in mind. (If you think my setup is a bad idea, please let me know & why. I also hope you can suggest a better one...) I am aware that I can only have up to 4 primary partitions, or 3 primary partitions with 1 extended parition max. Now, does the System Recovery portion count as one primary partition? I'm really new to these things and it is totally unclear to me. In shrinking my D: drive using Windows 7's Disk Management tool, I would get an unallocated free space which I don't know how to make an extended partition from. It seems like I can only create a primary partition from it, not an extended one. How do I go about it? (I'd also like to note, if it is of any importance, that I am trying to avoid using the option to install Ubuntu alongside Windows, and much rather prefer using the custom install where I can specify which drives I wish to use and stuff. Somehow I feel its safer that way.)

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  • How to migrate WinXP from failing old HD to new one

    - by Péter Török
    Following this issue, we have all our important data backed up now. I also bought and installed a new replacement hard disk (WD 160GB PATA) as secondary (slave) drive. I created two primary NTFS partitions on it: a 40 GB system partition, and a 110GB data partition. In theory I could start reinstalling WinXP from scratch on the new system partition, then copying over all user data from the old drive to the new data partition. Once this is done, I could even throw away the old drive, or keep it just to see what happens. (Note: I don't want to clone the whole drive as it contains a dual boot setup with an old Linux installation which I don't need anymore, and anyway, a fresh reinstall would do WinXP good to get rid of many years' clutter.) However, I am lazy :-) The old HD is still functioning, the problem has not manifested again since. So I feel there is no need to hurry with a complete OS reinstall. What I don't know though is whether I will be able to install WinXP on the new system partition at a later stage without affecting the contents of the data partition on the same drive. If this is possible, I can just move over all our data to the new data partition to have it safe, then continue running WinXP from the old drive as long as it works. Does anyone see any problems/risks with this plan?

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  • How to setup RAID 1 with Intel RST on an existing Windows 7 system?

    - by instcode
    I'd like to setup RAID-1 using Intel Rapid Storage Technology on my Windows 7 64-bit system. I have an 1TB SATA HDD with Windows 7 system installed on the first primary partition (leftmost, ~200GB). The rest of this HDD is unallocated (~800GB). I bought another 2TB SATA, then created a primary partition (leftmost, ~500GB) and filled my data in. The rest of this HDD is unallocated (~1.5TB). A quick disk layout (XXX is the unallocated region): HDD1 (1TB): [ 200GB C:\ SYSTEM | XXXXXXXXXXXX ] HDD2 (2TB): [ 500GB Z:\ PROGRAM | XXXXXXXXXXXXXXXXXXXXXX ] Now, I want to create a 500GB RAID-1 partition (I'm not sure if using "partition" is correct here) on the rightmost of the 2 HDDs above without losing any existing data from both disks. Here is the expected layout: HDD1 (1TB): [ 200GB C:\ SYSTEM | XXXXXX | 500GB D:\ DATA - RAID-1 ] HDD2 (2TB): [ 500GB Z:\ PROGRAM | XXXXXXXXXXXXXXXX | 500GB D:\ DATA RAID-1] Let's not concern about data lost, is it possible to have that final layout using Intel RST? Previously, I tried this layout using dynamic disk & software RAID from Windows and it worked as expected, however, it's quite ugly in resynching after an OS failure that I don't want. If yes, is there a way to keep the data on existing partitions untouched or, at least, it should keep the SYSTEM partition safe (I'm okay if the PROGRAM partition has to be gone.)? Well, are there any strict/special steps I should follow when using the Intel RST manager in order to achieve that? If none of those questions above are "Yes", could you please suggest some other possible layouts that leave the C:SYSTEM partition untouched?

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  • Hibernation fails; The system cannot find the file specified

    - by GMMan
    Recently I installed Ubuntu 12.04.1 LTS on my Lenovo Y480. Hibernation was working properly after the Ubuntu install, but I was making sure all of the operating systems on my system worked, including OneKey Recovery (recovery partition). It is of note that I installed Windows 7 from scratch with a disk image I downloaded off of my university's DreamSpark program, and further to that I had to image the partition with Paragon Backup & Recovery, repartition to convert the Windows partition to extended, install Ubuntu, and then restore the image. During that process I also used the Windows disc to edit the BCD as to reuse the existing entry for the restored partition. I also used the automated "repair your computer" option. With verification, I noticed that the "repair your computer" option actually wrote to the wrong BCD (the recovery partition), and I mounted the partition and restored the original BCD (from a copy I made earlier), and rebooted. At this point my GRUB broke, and I was able to restore it. At this point hibernation broke. I tried powercfg /h off and powercfg /h on, rebooted, and nothing. Also tried increasing the hibernation file size as directed on this post, but it still doesn't work. Executing shutdown /h yields The system cannot find the file specified.(2). What file? It seems that mounting the system partition sometimes works, but I don't want to keep it mounted in case it gets written to accidentally. How do I permanently fix this?

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  • Clone a Windows Installation to a 3TB Hard Drive; MBR to GPT

    - by DanBlakemore
    I have Windows 7 Professional 64-bit installed on my desktop. Unfortunately for me and my wallet my hard drive is failing. I have purchased a 3TB hard drive as a replacement for my current 2TB drive. I would like to avoid as much hassle as possible in moving to this new drive so I would like to copy my current partition to the new drive using Gparted. The problem is that I suspect that my current partition is MBR, and I need GPT on my new drive since it is 3TB. Can I simply copy the MBR partition onto the new disk and then convert it to GPT after the fact (can you even convert the type of a partition)? Or would I need to somehow copy the contents of the partition into a GPT partition on the new drive? How do I go about making this transistion? Also, are there any issues I should be wary of booting to a GPT partition? If it matters, my motherboard is 1 year old as of May, 2012. Edit: My motherboard is 1 day old. My old one does not have UEFI compatibility, so I decided to make an upgrade to Intel today given that I would need a UEFI motherboard to use my new HDD. How much can I use a dying hard drive (bad sectors according to Hitachi Drive Fitness Test)? I have assumed not at all, to be safe.

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  • Ubuntu 12.04 can't boot after installing with software RAID 1

    - by Bill
    I've been trying to install Ubuntu with software RAID on my server and there is obviously something that I don't understand about the process. This is the guide that I followed: https://help.ubuntu.com/11.04/serverguide/advanced-installation.html I have two identical 1 TB disks in my server. I went through the initial install process and manually set up my partitions. On each disk I set up: (1) 100 MB partition for EFI boot (I didn't originally have this but added it based on a forum post I found after my original install failed to boot, I ended up with EFIboot since that was what the 'guided partitioning' decided to do) (1) 970 MB partition for / (1) 30 MB partition for swap I then created new RAID 1 disks combining the two partitions, one from each disk, such that each partition is mirrored. I then configured their usage as stated above. After saving the configuration I said yes to boot in a degraded state. The rest of the setup went normally, no errors of any kind. I saw GRUB being installed and again no errors. However, after rebooting the server I get the dreaded 'Insert boot media' and nothing happens. I loaded up the recovery disk and the mdadm configuration looks correct. md0 is my EFIBoot partition md1 is my \ partition using ext4 md2 is my swap partition Running file -s /dev/md0 doesn't indicate that GRUB is there and so I attempted to reinstall GRUB using the recovery disk. I selected the md0 disk and it appeared to install just fine. Running file -s /dev/md1 shows the error needs journal recovery, I'm not sure if that's related or not or how to fix that. Rebooting gives me the same problem, no boot media found. I've searched around the internet but can't figure out what to do next or more importantly how to troubleshoot what exactly is going wrong. Thanks!

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  • How can I recover my data from a damaged hard drive?

    - by krk
    a few days ago when I was working on windows my laptop was beaten on the side where the hard drive is located. As a result, it was damaged and I couldn't access the windows partition. I had to boot the linux one, which is working without any trouble. I have 2 partition formatted with ntfs, the one with windows on it, and the other one intended to store data. I mounted the windows partition from ubuntu and I could see all my files. But when I tried to mount the data partition it was impossible. It threw me an error, it couldn't recognize ntfs partition. I try to copy the damaged disk into an external hard drive using the command: dd if=/dev/sda of=/dev/sdb conv=noerror,sync The progress stopped at 60%. I was still unable to mount the data partition. Now I'm trying to backup my files using an utility called Photorec. The problem is that it is recovering my files in a disorderly way, it is all mixed up and I need my original directory structure, it will become an endless task to organize the files as they were before. Is there any way I can get my partition back?

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  • Analytic functions – they’re not aggregates

    - by Rob Farley
    SQL 2012 brings us a bunch of new analytic functions, together with enhancements to the OVER clause. People who have known me over the years will remember that I’m a big fan of the OVER clause and the types of things that it brings us when applied to aggregate functions, as well as the ranking functions that it enables. The OVER clause was introduced in SQL Server 2005, and remained frustratingly unchanged until SQL Server 2012. This post is going to look at a particular aspect of the analytic functions though (not the enhancements to the OVER clause). When I give presentations about the analytic functions around Australia as part of the tour of SQL Saturdays (starting in Brisbane this Thursday), and in Chicago next month, I’ll make sure it’s sufficiently well described. But for this post – I’m going to skip that and assume you get it. The analytic functions introduced in SQL 2012 seem to come in pairs – FIRST_VALUE and LAST_VALUE, LAG and LEAD, CUME_DIST and PERCENT_RANK, PERCENTILE_CONT and PERCENTILE_DISC. Perhaps frustratingly, they take slightly different forms as well. The ones I want to look at now are FIRST_VALUE and LAST_VALUE, and PERCENTILE_CONT and PERCENTILE_DISC. The reason I’m pulling this ones out is that they always produce the same result within their partitions (if you’re applying them to the whole partition). Consider the following query: SELECT     YEAR(OrderDate),     FIRST_VALUE(TotalDue)         OVER (PARTITION BY YEAR(OrderDate)               ORDER BY OrderDate, SalesOrderID               RANGE BETWEEN UNBOUNDED PRECEDING                         AND UNBOUNDED FOLLOWING),     LAST_VALUE(TotalDue)         OVER (PARTITION BY YEAR(OrderDate)               ORDER BY OrderDate, SalesOrderID               RANGE BETWEEN UNBOUNDED PRECEDING                         AND UNBOUNDED FOLLOWING),     PERCENTILE_CONT(0.95)         WITHIN GROUP (ORDER BY TotalDue)         OVER (PARTITION BY YEAR(OrderDate)),     PERCENTILE_DISC(0.95)         WITHIN GROUP (ORDER BY TotalDue)         OVER (PARTITION BY YEAR(OrderDate)) FROM Sales.SalesOrderHeader ; This is designed to get the TotalDue for the first order of the year, the last order of the year, and also the 95% percentile, using both the continuous and discrete methods (‘discrete’ means it picks the closest one from the values available – ‘continuous’ means it will happily use something between, similar to what you would do for a traditional median of four values). I’m sure you can imagine the results – a different value for each field, but within each year, all the rows the same. Notice that I’m not grouping by the year. Nor am I filtering. This query gives us a result for every row in the SalesOrderHeader table – 31465 in this case (using the original AdventureWorks that dates back to the SQL 2005 days). The RANGE BETWEEN bit in FIRST_VALUE and LAST_VALUE is needed to make sure that we’re considering all the rows available. If we don’t specify that, it assumes we only mean “RANGE BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW”, which means that LAST_VALUE ends up being the row we’re looking at. At this point you might think about other environments such as Access or Reporting Services, and remember aggregate functions like FIRST. We really should be able to do something like: SELECT     YEAR(OrderDate),     FIRST_VALUE(TotalDue)         OVER (PARTITION BY YEAR(OrderDate)               ORDER BY OrderDate, SalesOrderID               RANGE BETWEEN UNBOUNDED PRECEDING                         AND UNBOUNDED FOLLOWING) FROM Sales.SalesOrderHeader GROUP BY YEAR(OrderDate) ; But you can’t. You get that age-old error: Msg 8120, Level 16, State 1, Line 5 Column 'Sales.SalesOrderHeader.OrderDate' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause. Msg 8120, Level 16, State 1, Line 5 Column 'Sales.SalesOrderHeader.SalesOrderID' is invalid in the select list because it is not contained in either an aggregate function or the GROUP BY clause. Hmm. You see, FIRST_VALUE isn’t an aggregate function. None of these analytic functions are. There are too many things involved for SQL to realise that the values produced might be identical within the group. Furthermore, you can’t even surround it in a MAX. Then you get a different error, telling you that you can’t use windowed functions in the context of an aggregate. And so we end up grouping by doing a DISTINCT. SELECT DISTINCT     YEAR(OrderDate),         FIRST_VALUE(TotalDue)              OVER (PARTITION BY YEAR(OrderDate)                   ORDER BY OrderDate, SalesOrderID                   RANGE BETWEEN UNBOUNDED PRECEDING                             AND UNBOUNDED FOLLOWING),         LAST_VALUE(TotalDue)             OVER (PARTITION BY YEAR(OrderDate)                   ORDER BY OrderDate, SalesOrderID                   RANGE BETWEEN UNBOUNDED PRECEDING                             AND UNBOUNDED FOLLOWING),     PERCENTILE_CONT(0.95)          WITHIN GROUP (ORDER BY TotalDue)         OVER (PARTITION BY YEAR(OrderDate)),     PERCENTILE_DISC(0.95)         WITHIN GROUP (ORDER BY TotalDue)         OVER (PARTITION BY YEAR(OrderDate)) FROM Sales.SalesOrderHeader ; I’m sorry. It’s just the way it goes. Hopefully it’ll change the future, but for now, it’s what you’ll have to do. If we look in the execution plan, we see that it’s incredibly ugly, and actually works out the results of these analytic functions for all 31465 rows, finally performing the distinct operation to convert it into the four rows we get in the results. You might be able to achieve a better plan using things like TOP, or the kind of calculation that I used in http://sqlblog.com/blogs/rob_farley/archive/2011/08/23/t-sql-thoughts-about-the-95th-percentile.aspx (which is how PERCENTILE_CONT works), but it’s definitely convenient to use these functions, and in time, I’m sure we’ll see good improvements in the way that they are implemented. Oh, and this post should be good for fellow SQL Server MVP Nigel Sammy’s T-SQL Tuesday this month.

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  • Fun with Aggregates

    - by Paul White
    There are interesting things to be learned from even the simplest queries.  For example, imagine you are given the task of writing a query to list AdventureWorks product names where the product has at least one entry in the transaction history table, but fewer than ten. One possible query to meet that specification is: SELECT p.Name FROM Production.Product AS p JOIN Production.TransactionHistory AS th ON p.ProductID = th.ProductID GROUP BY p.ProductID, p.Name HAVING COUNT_BIG(*) < 10; That query correctly returns 23 rows (execution plan and data sample shown below): The execution plan looks a bit different from the written form of the query: the base tables are accessed in reverse order, and the aggregation is performed before the join.  The general idea is to read all rows from the history table, compute the count of rows grouped by ProductID, merge join the results to the Product table on ProductID, and finally filter to only return rows where the count is less than ten. This ‘fully-optimized’ plan has an estimated cost of around 0.33 units.  The reason for the quote marks there is that this plan is not quite as optimal as it could be – surely it would make sense to push the Filter down past the join too?  To answer that, let’s look at some other ways to formulate this query.  This being SQL, there are any number of ways to write logically-equivalent query specifications, so we’ll just look at a couple of interesting ones.  The first query is an attempt to reverse-engineer T-SQL from the optimized query plan shown above.  It joins the result of pre-aggregating the history table to the Product table before filtering: SELECT p.Name FROM ( SELECT th.ProductID, cnt = COUNT_BIG(*) FROM Production.TransactionHistory AS th GROUP BY th.ProductID ) AS q1 JOIN Production.Product AS p ON p.ProductID = q1.ProductID WHERE q1.cnt < 10; Perhaps a little surprisingly, we get a slightly different execution plan: The results are the same (23 rows) but this time the Filter is pushed below the join!  The optimizer chooses nested loops for the join, because the cardinality estimate for rows passing the Filter is a bit low (estimate 1 versus 23 actual), though you can force a merge join with a hint and the Filter still appears below the join.  In yet another variation, the < 10 predicate can be ‘manually pushed’ by specifying it in a HAVING clause in the “q1” sub-query instead of in the WHERE clause as written above. The reason this predicate can be pushed past the join in this query form, but not in the original formulation is simply an optimizer limitation – it does make efforts (primarily during the simplification phase) to encourage logically-equivalent query specifications to produce the same execution plan, but the implementation is not completely comprehensive. Moving on to a second example, the following query specification results from phrasing the requirement as “list the products where there exists fewer than ten correlated rows in the history table”: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) < 10 ); Unfortunately, this query produces an incorrect result (86 rows): The problem is that it lists products with no history rows, though the reasons are interesting.  The COUNT_BIG(*) in the EXISTS clause is a scalar aggregate (meaning there is no GROUP BY clause) and scalar aggregates always produce a value, even when the input is an empty set.  In the case of the COUNT aggregate, the result of aggregating the empty set is zero (the other standard aggregates produce a NULL).  To make the point really clear, let’s look at product 709, which happens to be one for which no history rows exist: -- Scalar aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709;   -- Vector aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709 GROUP BY th.ProductID; The estimated execution plans for these two statements are almost identical: You might expect the Stream Aggregate to have a Group By for the second statement, but this is not the case.  The query includes an equality comparison to a constant value (709), so all qualified rows are guaranteed to have the same value for ProductID and the Group By is optimized away. In fact there are some minor differences between the two plans (the first is auto-parameterized and qualifies for trivial plan, whereas the second is not auto-parameterized and requires cost-based optimization), but there is nothing to indicate that one is a scalar aggregate and the other is a vector aggregate.  This is something I would like to see exposed in show plan so I suggested it on Connect.  Anyway, the results of running the two queries show the difference at runtime: The scalar aggregate (no GROUP BY) returns a result of zero, whereas the vector aggregate (with a GROUP BY clause) returns nothing at all.  Returning to our EXISTS query, we could ‘fix’ it by changing the HAVING clause to reject rows where the scalar aggregate returns zero: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) BETWEEN 1 AND 9 ); The query now returns the correct 23 rows: Unfortunately, the execution plan is less efficient now – it has an estimated cost of 0.78 compared to 0.33 for the earlier plans.  Let’s try adding a redundant GROUP BY instead of changing the HAVING clause: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY th.ProductID HAVING COUNT_BIG(*) < 10 ); Not only do we now get correct results (23 rows), this is the execution plan: I like to compare that plan to quantum physics: if you don’t find it shocking, you haven’t understood it properly :)  The simple addition of a redundant GROUP BY has resulted in the EXISTS form of the query being transformed into exactly the same optimal plan we found earlier.  What’s more, in SQL Server 2008 and later, we can replace the odd-looking GROUP BY with an explicit GROUP BY on the empty set: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ); I offer that as an alternative because some people find it more intuitive (and it perhaps has more geek value too).  Whichever way you prefer, it’s rather satisfying to note that the result of the sub-query does not exist for a particular correlated value where a vector aggregate is used (the scalar COUNT aggregate always returns a value, even if zero, so it always ‘EXISTS’ regardless which ProductID is logically being evaluated). The following query forms also produce the optimal plan and correct results, so long as a vector aggregate is used (you can probably find more equivalent query forms): WHERE Clause SELECT p.Name FROM Production.Product AS p WHERE ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) < 10; APPLY SELECT p.Name FROM Production.Product AS p CROSS APPLY ( SELECT NULL FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ) AS ca (dummy); FROM Clause SELECT q1.Name FROM ( SELECT p.Name, cnt = ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) FROM Production.Product AS p ) AS q1 WHERE q1.cnt < 10; This last example uses SUM(1) instead of COUNT and does not require a vector aggregate…you should be able to work out why :) SELECT q.Name FROM ( SELECT p.Name, cnt = ( SELECT SUM(1) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID ) FROM Production.Product AS p ) AS q WHERE q.cnt < 10; The semantics of SQL aggregates are rather odd in places.  It definitely pays to get to know the rules, and to be careful to check whether your queries are using scalar or vector aggregates.  As we have seen, query plans do not show in which ‘mode’ an aggregate is running and getting it wrong can cause poor performance, wrong results, or both. © 2012 Paul White Twitter: @SQL_Kiwi email: [email protected]

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  • Fun with Aggregates

    - by Paul White
    There are interesting things to be learned from even the simplest queries.  For example, imagine you are given the task of writing a query to list AdventureWorks product names where the product has at least one entry in the transaction history table, but fewer than ten. One possible query to meet that specification is: SELECT p.Name FROM Production.Product AS p JOIN Production.TransactionHistory AS th ON p.ProductID = th.ProductID GROUP BY p.ProductID, p.Name HAVING COUNT_BIG(*) < 10; That query correctly returns 23 rows (execution plan and data sample shown below): The execution plan looks a bit different from the written form of the query: the base tables are accessed in reverse order, and the aggregation is performed before the join.  The general idea is to read all rows from the history table, compute the count of rows grouped by ProductID, merge join the results to the Product table on ProductID, and finally filter to only return rows where the count is less than ten. This ‘fully-optimized’ plan has an estimated cost of around 0.33 units.  The reason for the quote marks there is that this plan is not quite as optimal as it could be – surely it would make sense to push the Filter down past the join too?  To answer that, let’s look at some other ways to formulate this query.  This being SQL, there are any number of ways to write logically-equivalent query specifications, so we’ll just look at a couple of interesting ones.  The first query is an attempt to reverse-engineer T-SQL from the optimized query plan shown above.  It joins the result of pre-aggregating the history table to the Product table before filtering: SELECT p.Name FROM ( SELECT th.ProductID, cnt = COUNT_BIG(*) FROM Production.TransactionHistory AS th GROUP BY th.ProductID ) AS q1 JOIN Production.Product AS p ON p.ProductID = q1.ProductID WHERE q1.cnt < 10; Perhaps a little surprisingly, we get a slightly different execution plan: The results are the same (23 rows) but this time the Filter is pushed below the join!  The optimizer chooses nested loops for the join, because the cardinality estimate for rows passing the Filter is a bit low (estimate 1 versus 23 actual), though you can force a merge join with a hint and the Filter still appears below the join.  In yet another variation, the < 10 predicate can be ‘manually pushed’ by specifying it in a HAVING clause in the “q1” sub-query instead of in the WHERE clause as written above. The reason this predicate can be pushed past the join in this query form, but not in the original formulation is simply an optimizer limitation – it does make efforts (primarily during the simplification phase) to encourage logically-equivalent query specifications to produce the same execution plan, but the implementation is not completely comprehensive. Moving on to a second example, the following query specification results from phrasing the requirement as “list the products where there exists fewer than ten correlated rows in the history table”: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) < 10 ); Unfortunately, this query produces an incorrect result (86 rows): The problem is that it lists products with no history rows, though the reasons are interesting.  The COUNT_BIG(*) in the EXISTS clause is a scalar aggregate (meaning there is no GROUP BY clause) and scalar aggregates always produce a value, even when the input is an empty set.  In the case of the COUNT aggregate, the result of aggregating the empty set is zero (the other standard aggregates produce a NULL).  To make the point really clear, let’s look at product 709, which happens to be one for which no history rows exist: -- Scalar aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709;   -- Vector aggregate SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = 709 GROUP BY th.ProductID; The estimated execution plans for these two statements are almost identical: You might expect the Stream Aggregate to have a Group By for the second statement, but this is not the case.  The query includes an equality comparison to a constant value (709), so all qualified rows are guaranteed to have the same value for ProductID and the Group By is optimized away. In fact there are some minor differences between the two plans (the first is auto-parameterized and qualifies for trivial plan, whereas the second is not auto-parameterized and requires cost-based optimization), but there is nothing to indicate that one is a scalar aggregate and the other is a vector aggregate.  This is something I would like to see exposed in show plan so I suggested it on Connect.  Anyway, the results of running the two queries show the difference at runtime: The scalar aggregate (no GROUP BY) returns a result of zero, whereas the vector aggregate (with a GROUP BY clause) returns nothing at all.  Returning to our EXISTS query, we could ‘fix’ it by changing the HAVING clause to reject rows where the scalar aggregate returns zero: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID HAVING COUNT_BIG(*) BETWEEN 1 AND 9 ); The query now returns the correct 23 rows: Unfortunately, the execution plan is less efficient now – it has an estimated cost of 0.78 compared to 0.33 for the earlier plans.  Let’s try adding a redundant GROUP BY instead of changing the HAVING clause: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY th.ProductID HAVING COUNT_BIG(*) < 10 ); Not only do we now get correct results (23 rows), this is the execution plan: I like to compare that plan to quantum physics: if you don’t find it shocking, you haven’t understood it properly :)  The simple addition of a redundant GROUP BY has resulted in the EXISTS form of the query being transformed into exactly the same optimal plan we found earlier.  What’s more, in SQL Server 2008 and later, we can replace the odd-looking GROUP BY with an explicit GROUP BY on the empty set: SELECT p.Name FROM Production.Product AS p WHERE EXISTS ( SELECT * FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ); I offer that as an alternative because some people find it more intuitive (and it perhaps has more geek value too).  Whichever way you prefer, it’s rather satisfying to note that the result of the sub-query does not exist for a particular correlated value where a vector aggregate is used (the scalar COUNT aggregate always returns a value, even if zero, so it always ‘EXISTS’ regardless which ProductID is logically being evaluated). The following query forms also produce the optimal plan and correct results, so long as a vector aggregate is used (you can probably find more equivalent query forms): WHERE Clause SELECT p.Name FROM Production.Product AS p WHERE ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) < 10; APPLY SELECT p.Name FROM Production.Product AS p CROSS APPLY ( SELECT NULL FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () HAVING COUNT_BIG(*) < 10 ) AS ca (dummy); FROM Clause SELECT q1.Name FROM ( SELECT p.Name, cnt = ( SELECT COUNT_BIG(*) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID GROUP BY () ) FROM Production.Product AS p ) AS q1 WHERE q1.cnt < 10; This last example uses SUM(1) instead of COUNT and does not require a vector aggregate…you should be able to work out why :) SELECT q.Name FROM ( SELECT p.Name, cnt = ( SELECT SUM(1) FROM Production.TransactionHistory AS th WHERE th.ProductID = p.ProductID ) FROM Production.Product AS p ) AS q WHERE q.cnt < 10; The semantics of SQL aggregates are rather odd in places.  It definitely pays to get to know the rules, and to be careful to check whether your queries are using scalar or vector aggregates.  As we have seen, query plans do not show in which ‘mode’ an aggregate is running and getting it wrong can cause poor performance, wrong results, or both. © 2012 Paul White Twitter: @SQL_Kiwi email: [email protected]

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  • Can't install Ubuntu 11.10 on a seperate disk partition. I have a log file. Please help!

    - by les02jen17
    Before I installed Ubuntu, I resized my C: drive (which has my Windows 7 OS). Prior to resizing, I even defragged the said drive. I tried using Wubi to install it on the created partition, but I receive an error. I have the log file and I would post it but it's too long and the site won't let me. :(So anyway, I wasn't able to install it, so I uninstalled it. Upon uninstalling Wubi and Ubuntu, I tried installing it by having the CD to my CD ROM. The installation was a success, however, upon rebooting I can neither boot to Ubuntu OR Windows 7. I get stuck in an infinite bootloop. I hope you can help me out! :(

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  • SQL SERVER – Weekly Series – Memory Lane – #037

    - by Pinal Dave
    Here is the list of selected articles of SQLAuthority.com across all these years. Instead of just listing all the articles I have selected a few of my most favorite articles and have listed them here with additional notes below it. Let me know which one of the following is your favorite article from memory lane. 2007 Convert Text to Numbers (Integer) – CAST and CONVERT If table column is VARCHAR and has all the numeric values in it, it can be retrieved as Integer using CAST or CONVERT function. List All Stored Procedure Modified in Last N Days If SQL Server suddenly start behaving in un-expectable behavior and if stored procedure were changed recently, following script can be used to check recently modified stored procedure. If a stored procedure was created but never modified afterwards modified date and create a date for that stored procedure are same. Count Duplicate Records – Rows Validate Field For DATE datatype using function ISDATE() We always checked DATETIME field for incorrect data type. One of the user input date as 30/2/2007. The date was sucessfully inserted in the temp table but while inserting from temp table to final table it crashed with error. We had now task to validate incorrect date value before we insert in final table. Jr. Developer asked me how can he do that? We check for incorrect data type (varchar, int, NULL) but this is incorrect date value. Regular expression works fine with them because of mm/dd/yyyy format. 2008 Find Space Used For Any Particular Table It is very simple to find out the space used by any table in the database. Two Convenient Features Inline Assignment – Inline Operations Here is the script which does both – Inline Assignment and Inline Operation DECLARE @idx INT = 0 SET @idx+=1 SELECT @idx Introduction to SPARSE Columns SPARSE column are better at managing NULL and ZERO values in SQL Server. It does not take any space in database at all. If column is created with SPARSE clause with it and it contains ZERO or NULL it will be take lesser space then regular column (without SPARSE clause). SP_CONFIGURE – Displays or Changes Global Configuration Settings If advanced settings are not enabled at configuration level SQL Server will not let user change the advanced features on server. Authorized user can turn on or turn off advance settings. 2009 Standby Servers and Types of Standby Servers Standby Server is a type of server that can be brought online in a situation when Primary Server goes offline and application needs continuous (high) availability of the server. There is always a need to set up a mechanism where data and objects from primary server are moved to secondary (standby) server. BLOB – Pointer to Image, Image in Database, FILESTREAM Storage When it comes to storing images in database there are two common methods. I had previously blogged about the same subject on my visit to Toronto. With SQL Server 2008, we have a new method of FILESTREAM storage. However, the answer on when to use FILESTREAM and when to use other methods is still vague in community. 2010 Upper Case Shortcut SQL Server Management Studio I select the word and hit CTRL+SHIFT+U and it SSMS immediately changes the case of the selected word. Similar way if one want to convert cases to lower case, another short cut CTRL+SHIFT+L is also available. The Self Join – Inner Join and Outer Join Self Join has always been a noteworthy case. It is interesting to ask questions about self join in a room full of developers. I often ask – if there are three kinds of joins, i.e.- Inner Join, Outer Join and Cross Join; what type of join is Self Join? The usual answer is that it is an Inner Join. However, the reality is very different. Parallelism – Row per Processor – Row per Thread – Thread 0  If you look carefully in the Properties window or XML Plan, there is “Thread 0?. What does this “Thread 0” indicate? Well find out from the blog post. How do I Learn and How do I Teach The blog post has raised three very interesting questions. How do you learn? How do you teach? What are you learning or teaching? Let me try to answer the same. 2011 SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 7 of 31 What are Different Types of Locks? What are Pessimistic Lock and Optimistic Lock? When is the use of UPDATE_STATISTICS command? What is the Difference between a HAVING clause and a WHERE clause? What is Connection Pooling and why it is Used? What are the Properties and Different Types of Sub-Queries? What are the Authentication Modes in SQL Server? How can it be Changed? SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 8 of 31 Which Command using Query Analyzer will give you the Version of SQL Server and Operating System? What is an SQL Server Agent? Can a Stored Procedure call itself or a Recursive Stored Procedure? How many levels of SP nesting is possible? What is Log Shipping? Name 3 ways to get an Accurate Count of the Number of Records in a Table? What does it mean to have QUOTED_IDENTIFIER ON? What are the Implications of having it OFF? What is the Difference between a Local and a Global Temporary Table? What is the STUFF Function and How Does it Differ from the REPLACE Function? What is PRIMARY KEY? What is UNIQUE KEY Constraint? What is FOREIGN KEY? SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 9 of 31 What is CHECK Constraint? What is NOT NULL Constraint? What is the difference between UNION and UNION ALL? What is B-Tree? How to get @@ERROR and @@ROWCOUNT at the Same Time? What is a Scheduled Job or What is a Scheduled Task? What are the Advantages of Using Stored Procedures? What is a Table Called, if it has neither Cluster nor Non-cluster Index? What is it Used for? Can SQL Servers Linked to other Servers like Oracle? What is BCP? When is it Used? SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 10 of 31 What Command do we Use to Rename a db, a Table and a Column? What are sp_configure Commands and SET Commands? How to Implement One-to-One, One-to-Many and Many-to-Many Relationships while Designing Tables? What is Difference between Commit and Rollback when Used in Transactions? What is an Execution Plan? When would you Use it? How would you View the Execution Plan? SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 11 of 31 What is Difference between Table Aliases and Column Aliases? Do they Affect Performance? What is the difference between CHAR and VARCHAR Datatypes? What is the Difference between VARCHAR and VARCHAR(MAX) Datatypes? What is the Difference between VARCHAR and NVARCHAR datatypes? Which are the Important Points to Note when Multilanguage Data is Stored in a Table? How to Optimize Stored Procedure Optimization? What is SQL Injection? How to Protect Against SQL Injection Attack? How to Find Out the List Schema Name and Table Name for the Database? What is CHECKPOINT Process in the SQL Server? SQL SERVER – Interview Questions and Answers – Frequently Asked Questions – Day 12 of 31 How does Using a Separate Hard Drive for Several Database Objects Improves Performance Right Away? How to Find the List of Fixed Hard Drive and Free Space on Server? Why can there be only one Clustered Index and not more than one? What is Difference between Line Feed (\n) and Carriage Return (\r)? Is It Possible to have Clustered Index on Separate Drive From Original Table Location? What is a Hint? How to Delete Duplicate Rows? Why the Trigger Fires Multiple Times in Single Login? 2012 CTRL+SHIFT+] Shortcut to Select Code Between Two Parenthesis Shortcut key is CTRL+SHIFT+]. This key can be very useful when dealing with multiple subqueries, CTE or query with multiple parentheses. When exercised this shortcut key it selects T-SQL code between two parentheses. Monday Morning Puzzle – Query Returns Results Sometimes but Not Always I am beginner with SQL Server. I have one query, it sometime returns a result and sometime it does not return me the result. Where should I start looking for a solution and what kind of information I should send to you so you can help me with solving. I have no clue, please guide me. Remove Debug Button in SSMS – SQL in Sixty Seconds #020 – Video Effect of Case Sensitive Collation on Resultset Collation is a very interesting concept but I quite often see it is heavily neglected. I have seen developer and DBA looking for a workaround to fix collation error rather than understanding if the side effect of the workaround. Switch Between Two Parenthesis using Shortcut CTRL+] Earlier this week I wrote a blog post about CTRL+SHIFT+] Shortcut to Select Code Between Two Parenthesis, I received quite a lot of positive feedback from readers. If you are a regular reader of the blog post, you must be aware that I appreciate the learning shared by readers. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Memory Lane, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • More CPU cores may not always lead to better performance – MAXDOP and query memory distribution in spotlight

    - by sqlworkshops
    More hardware normally delivers better performance, but there are exceptions where it can hinder performance. Understanding these exceptions and working around it is a major part of SQL Server performance tuning.   When a memory allocating query executes in parallel, SQL Server distributes memory to each task that is executing part of the query in parallel. In our example the sort operator that executes in parallel divides the memory across all tasks assuming even distribution of rows. Common memory allocating queries are that perform Sort and do Hash Match operations like Hash Join or Hash Aggregation or Hash Union.   In reality, how often are column values evenly distributed, think about an example; are employees working for your company distributed evenly across all the Zip codes or mainly concentrated in the headquarters? What happens when you sort result set based on Zip codes? Do all products in the catalog sell equally or are few products hot selling items?   One of my customers tested the below example on a 24 core server with various MAXDOP settings and here are the results:MAXDOP 1: CPU time = 1185 ms, elapsed time = 1188 msMAXDOP 4: CPU time = 1981 ms, elapsed time = 1568 msMAXDOP 8: CPU time = 1918 ms, elapsed time = 1619 msMAXDOP 12: CPU time = 2367 ms, elapsed time = 2258 msMAXDOP 16: CPU time = 2540 ms, elapsed time = 2579 msMAXDOP 20: CPU time = 2470 ms, elapsed time = 2534 msMAXDOP 0: CPU time = 2809 ms, elapsed time = 2721 ms - all 24 cores.In the above test, when the data was evenly distributed, the elapsed time of parallel query was always lower than serial query.   Why does the query get slower and slower with more CPU cores / higher MAXDOP? Maybe you can answer this question after reading the article; let me know: [email protected].   Well you get the point, let’s see an example.   The best way to learn is to practice. To create the below tables and reproduce the behavior, join the mailing list by using this link: www.sqlworkshops.com/ml and I will send you the table creation script.   Let’s update the Employees table with 49 out of 50 employees located in Zip code 2001. update Employees set Zip = EmployeeID / 400 + 1 where EmployeeID % 50 = 1 update Employees set Zip = 2001 where EmployeeID % 50 != 1 go update statistics Employees with fullscan go   Let’s create the temporary table #FireDrill with all possible Zip codes. drop table #FireDrill go create table #FireDrill (Zip int primary key) insert into #FireDrill select distinct Zip from Employees update statistics #FireDrill with fullscan go  Let’s execute the query serially with MAXDOP 1. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --First serially with MAXDOP 1 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 1) goThe query took 1011 ms to complete.   The execution plan shows the 77816 KB of memory was granted while the estimated rows were 799624.  No Sort Warnings in SQL Server Profiler.  Now let’s execute the query in parallel with MAXDOP 0. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --In parallel with MAXDOP 0 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 0) go The query took 1912 ms to complete.  The execution plan shows the 79360 KB of memory was granted while the estimated rows were 799624.  The estimated number of rows between serial and parallel plan are the same. The parallel plan has slightly more memory granted due to additional overhead. Sort properties shows the rows are unevenly distributed over the 4 threads.   Sort Warnings in SQL Server Profiler.   Intermediate Summary: The reason for the higher duration with parallel plan was sort spill. This is due to uneven distribution of employees over Zip codes, especially concentration of 49 out of 50 employees in Zip code 2001. Now let’s update the Employees table and distribute employees evenly across all Zip codes.   update Employees set Zip = EmployeeID / 400 + 1 go update statistics Employees with fullscan go  Let’s execute the query serially with MAXDOP 1. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --Serially with MAXDOP 1 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 1) go   The query took 751 ms to complete.  The execution plan shows the 77816 KB of memory was granted while the estimated rows were 784707.  No Sort Warnings in SQL Server Profiler.   Now let’s execute the query in parallel with MAXDOP 0. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --In parallel with MAXDOP 0 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 0) go The query took 661 ms to complete.  The execution plan shows the 79360 KB of memory was granted while the estimated rows were 784707.  Sort properties shows the rows are evenly distributed over the 4 threads. No Sort Warnings in SQL Server Profiler.    Intermediate Summary: When employees were distributed unevenly, concentrated on 1 Zip code, parallel sort spilled while serial sort performed well without spilling to tempdb. When the employees were distributed evenly across all Zip codes, parallel sort and serial sort did not spill to tempdb. This shows uneven data distribution may affect the performance of some parallel queries negatively. For detailed discussion of memory allocation, refer to webcasts available at www.sqlworkshops.com/webcasts.     Some of you might conclude from the above execution times that parallel query is not faster even when there is no spill. Below you can see when we are joining limited amount of Zip codes, parallel query will be fasted since it can use Bitmap Filtering.   Let’s update the Employees table with 49 out of 50 employees located in Zip code 2001. update Employees set Zip = EmployeeID / 400 + 1 where EmployeeID % 50 = 1 update Employees set Zip = 2001 where EmployeeID % 50 != 1 go update statistics Employees with fullscan go  Let’s create the temporary table #FireDrill with limited Zip codes. drop table #FireDrill go create table #FireDrill (Zip int primary key) insert into #FireDrill select distinct Zip       from Employees where Zip between 1800 and 2001 update statistics #FireDrill with fullscan go  Let’s execute the query serially with MAXDOP 1. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --Serially with MAXDOP 1 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 1) go The query took 989 ms to complete.  The execution plan shows the 77816 KB of memory was granted while the estimated rows were 785594. No Sort Warnings in SQL Server Profiler.  Now let’s execute the query in parallel with MAXDOP 0. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --In parallel with MAXDOP 0 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 0) go The query took 1799 ms to complete.  The execution plan shows the 79360 KB of memory was granted while the estimated rows were 785594.  Sort Warnings in SQL Server Profiler.    The estimated number of rows between serial and parallel plan are the same. The parallel plan has slightly more memory granted due to additional overhead.  Intermediate Summary: The reason for the higher duration with parallel plan even with limited amount of Zip codes was sort spill. This is due to uneven distribution of employees over Zip codes, especially concentration of 49 out of 50 employees in Zip code 2001.   Now let’s update the Employees table and distribute employees evenly across all Zip codes. update Employees set Zip = EmployeeID / 400 + 1 go update statistics Employees with fullscan go Let’s execute the query serially with MAXDOP 1. --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --Serially with MAXDOP 1 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 1) go The query took 250  ms to complete.  The execution plan shows the 9016 KB of memory was granted while the estimated rows were 79973.8.  No Sort Warnings in SQL Server Profiler.  Now let’s execute the query in parallel with MAXDOP 0.  --Example provided by www.sqlworkshops.com --Execute query with uneven Zip code distribution --In parallel with MAXDOP 0 set statistics time on go declare @EmployeeID int, @EmployeeName varchar(48),@zip int select @EmployeeName = e.EmployeeName, @zip = e.Zip from Employees e       inner join #FireDrill fd on (e.Zip = fd.Zip)       order by e.Zip option (maxdop 0) go The query took 85 ms to complete.  The execution plan shows the 13152 KB of memory was granted while the estimated rows were 784707.  No Sort Warnings in SQL Server Profiler.    Here you see, parallel query is much faster than serial query since SQL Server is using Bitmap Filtering to eliminate rows before the hash join.   Parallel queries are very good for performance, but in some cases it can hinder performance. If one identifies the reason for these hindrances, then it is possible to get the best out of parallelism. I covered many aspects of monitoring and tuning parallel queries in webcasts (www.sqlworkshops.com/webcasts) and articles (www.sqlworkshops.com/articles). I suggest you to watch the webcasts and read the articles to better understand how to identify and tune parallel query performance issues.   Summary: One has to avoid sort spill over tempdb and the chances of spills are higher when a query executes in parallel with uneven data distribution. Parallel query brings its own advantage, reduced elapsed time and reduced work with Bitmap Filtering. So it is important to understand how to avoid spills over tempdb and when to execute a query in parallel.   I explain these concepts with detailed examples in my webcasts (www.sqlworkshops.com/webcasts), I recommend you to watch them. The best way to learn is to practice. To create the above tables and reproduce the behavior, join the mailing list at www.sqlworkshops.com/ml and I will send you the relevant SQL Scripts.   Register for the upcoming 3 Day Level 400 Microsoft SQL Server 2008 and SQL Server 2005 Performance Monitoring & Tuning Hands-on Workshop in London, United Kingdom during March 15-17, 2011, click here to register / Microsoft UK TechNet.These are hands-on workshops with a maximum of 12 participants and not lectures. For consulting engagements click here.   Disclaimer and copyright information:This article refers to organizations and products that may be the trademarks or registered trademarks of their various owners. Copyright of this article belongs to R Meyyappan / www.sqlworkshops.com. You may freely use the ideas and concepts discussed in this article with acknowledgement (www.sqlworkshops.com), but you may not claim any of it as your own work. This article is for informational purposes only; you use any of the suggestions given here entirely at your own risk.   Register for the upcoming 3 Day Level 400 Microsoft SQL Server 2008 and SQL Server 2005 Performance Monitoring & Tuning Hands-on Workshop in London, United Kingdom during March 15-17, 2011, click here to register / Microsoft UK TechNet.These are hands-on workshops with a maximum of 12 participants and not lectures. For consulting engagements click here.   R Meyyappan [email protected] LinkedIn: http://at.linkedin.com/in/rmeyyappan  

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