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  • MySQL storage: how to manage a grow to infinite ?

    - by Dario
    Hi, I'm just thinking about famous internet services like facebook or twitter manage fast growing databases. Which could be a solution for this kind of problem? What about ids ? I read there is a limit in MySQL - 18446744073709551615 - in unsigned bigint... whow would you generate and manage a bigger value ? Just a theoric problem, but i'm curious about a possible solution. Thank you!

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  • How do I uninstall MySQL on Mac OS X (Snow Leopard)?

    - by Abhic
    I had installed MySQL from the command line when I setup a local web server with custom modules on my Snow Leopard box. I recently discovered MAMP and its just easier to work with it than with the personal web server Apple has and via command like utils. I uninstall ports completely but still see 'mysqld' in my Activity Monitor. I would like to uninstall this cleanly. Any tips?

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  • MySQL asking a user for SUPER privilege to perform a delete.

    - by Fran
    Hello, When trying to do a delete operation on a table, mysql reports the following error: Error code 1227: Access denied; you need the SUPER privilege for this operation. However, my user has this privilege granted for all tables in the schema: GRANT ALL PRIVILEGES ON myschema.* TO 'my_admin'@'%' How come it asks me for SUPER privilege for a delete? Thanks in advance.

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  • How Do I Reset My Root Password on MySQL on a Mac?

    - by classicrock985
    I'm using SequelPro (http://www.sequelpro.com/) and I would like to know how to reset my root password. I'm trying to log in as host: localhost username: root password: (BLANK) but I keep getting this error: Unable to connect to host because access was denied. Double-check your username and password and ensure that access from your current location is permitted. MySQL said: Access denied for user 'root'@'localhost' (using password: NO) Any suggestions?!

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  • What is the best way to secure MySQL data on a laptop *without* whole-disk-encryption?

    - by GJ
    I need to have the mysql data on my laptop stored in an encrypted state so that in case of the laptop being lost/stolen it will extremely difficult to recover the data without the password. I don't wish to use whole disk encryption, due to the performance impact it will have on other disk-intensive programs' usage. What could be the ideal solution for me balancing security and performance? Thanks!

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  • FTP upload for a PHP file hosting site, how to connect ProFTPD to mysql database?

    - by Igor
    I'm running a file upload service and users have requested to have FTP upload features Basically, I need to allow users to login, via FTP, to an FTP daemon (say, proFTPd) and they should be able to use their username and password (stored in a mysql database) to login there After logging in, I'll take care of the files with a cron job I'm stuck on how to make proftpd get users and passwords from my database..any ideas?

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  • No rule to make target libmysql.c', needed bylibmysql.lo'. Stop

    - by user1711008
    I install mysql5.1.53, run #./configure is well, but run #make have this error. My system is centos5.8, gcc version 4.1.2 20080704 (Red Hat 4.1.2-52) make[2]: Leaving directory /root/soft/mysql-5.1.53/libmysql' make[1]: Leaving directory/root/soft/mysql-5.1.53/libmysql' Making all in libmysql_r make[1]: Entering directory /root/soft/mysql-5.1.53/libmysql_r' make all-am make[2]: Entering directory/root/soft/mysql-5.1.53/libmysql_r' make[2]: * No rule to make target libmysql.c', needed bylibmysql.lo'. Stop. make[2]: Leaving directory /root/soft/mysql-5.1.53/libmysql_r' make[1]: *** [all] Error 2 make[1]: Leaving directory/root/soft/mysql-5.1.53/libmysql_r' make: * [all-recursive] Error 1

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  • What does this error mean (Can't create TCP/IP socket (24))?

    - by user105196
    I have web server with OS RHEL 6.2 and Mysql 5.5.23 on another server and the web server can read from Mysql server without problem, but some time I got this error: [Sun Sep 23 06:13:07 2012] [error] [client XXXXX] DBI connect('XXXX:192.168.1.2:3306','XXX',...) failed: Can't create TCP/IP socket (24) at /var/www/html/file.pm line 199. my question : What does this error mean (Can't create TCP/IP socket (24))? is it OS error or Mysql error ? perl -v This is perl, v5.10.1 (*) built for x86_64-linux-thread-multi mysql -V mysql Ver 14.14 Distrib 5.5.23, for Linux (x86_64) using readline 5.1 su - mysql -s /bin/bash -c 'ulimit -a' core file size (blocks, -c) 0 data seg size (kbytes, -d) unlimited scheduling priority (-e) 0 file size (blocks, -f) unlimited pending signals (-i) 127220 max locked memory (kbytes, -l) 64 max memory size (kbytes, -m) unlimited open files (-n) 1024 pipe size (512 bytes, -p) 8 POSIX message queues (bytes, -q) 819200 real-time priority (-r) 0 stack size (kbytes, -s) 10240 cpu time (seconds, -t) unlimited max user processes (-u) 1024 virtual memory (kbytes, -v) unlimited file locks (-x) unlimited

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  • How do I completely uninstall mySQL on XP, including the root password?

    - by user341219
    All I need is to be able to log in using root, but have forgotten the password. None of the steps to reset i found online work (i don't even have some of the executables mentioned such as mysql-nt.exe) However I have no problem deleting all databases (i have scripts) and intallations and starting completely from scratch... but uninstalling and deleting directories doesn't work. Thanks.

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  • SQL SERVER – Order By Numeric Values Formatted as String

    - by pinaldave
    When I was writing this blog post I had a hard time to come up with the title of the blog post so I did my best to come up with one. Here is the reason why? I wrote a blog post earlier SQL SERVER – Find First Non-Numeric Character from String. One of the questions was that how that blog can be useful in real life scenario. This blog post is the answer to that question. Let us first see a problem. We have a table which has a column containing alphanumeric data. The data always has first as an integer and later part as a string. The business need is to order the data based on the first part of the alphanumeric data which is an integer. Now the problem is that no matter how we use ORDER BY the result is not produced as expected. Let us understand this with example. Prepare a sample data: -- How to find first non numberic character USE tempdb GO CREATE TABLE MyTable (ID INT, Col1 VARCHAR(100)) GO INSERT INTO MyTable (ID, Col1) SELECT 1, '1one' UNION ALL SELECT 2, '11eleven' UNION ALL SELECT 3, '2two' UNION ALL SELECT 4, '22twentytwo' UNION ALL SELECT 5, '111oneeleven' GO -- Select Data SELECT * FROM MyTable GO The above query will give following result set. Now let us use ORDER BY COL1 and observe the result along with Original SELECT. -- Select Data SELECT * FROM MyTable GO -- Select Data SELECT * FROM MyTable ORDER BY Col1 GO The result of the table is not as per expected. We need the result in following format. Here is the good example of how we can use PATINDEX. -- Use of PATINDEX SELECT ID, LEFT(Col1,PATINDEX('%[^0-9]%',Col1)-1) 'Numeric Character', Col1 'Original Character' FROM MyTable ORDER BY LEFT(Col1,PATINDEX('%[^0-9]%',Col1)-1) GO We can use PATINDEX to identify the length of the digit part in the alphanumeric string (Remember: Our string has a first part as an int always. It will not work in any other scenario). Now you can use the LEFT function to extract the INT portion from the alphanumeric string and order the data according to it. You can easily clean up the script by dropping following table. DROP TABLE MyTable GO Here is the complete script so you can easily refer it. -- How to find first non numberic character USE tempdb GO CREATE TABLE MyTable (ID INT, Col1 VARCHAR(100)) GO INSERT INTO MyTable (ID, Col1) SELECT 1, '1one' UNION ALL SELECT 2, '11eleven' UNION ALL SELECT 3, '2two' UNION ALL SELECT 4, '22twentytwo' UNION ALL SELECT 5, '111oneeleven' GO -- Select Data SELECT * FROM MyTable GO -- Select Data SELECT * FROM MyTable ORDER BY Col1 GO -- Use of PATINDEX SELECT ID, Col1 'Original Character' FROM MyTable ORDER BY LEFT(Col1,PATINDEX('%[^0-9]%',Col1)-1) GO DROP TABLE MyTable GO Well, isn’t it an interesting solution. Any suggestion for better solution? Additionally any suggestion for changing the title of this blog post? Reference : Pinal Dave (http://blog.SQLAuthority.com) Filed under: PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL String, SQL Tips and Tricks, T SQL, Technology

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  • More useful Sql Server Serivce Broker Queries

    - by ChrisD
    SELECT 'Checking Broker Service Status...' IF (select Top 1 is_broker_enabled from sys.databases where name = 'NWMESSAGE')=1     SELECT ' Broker Service IS Enabled'  -- Should return a 1. ELSE     SELECT '** Broker Service IS DISABLED ***' /* If Is_Broker_enabled returns 0, uncomment and run this code ALTER DATABASE NWMESSAGE SET SINGLE_USER WITH ROLLBACK IMMEDIATE GO Alter Database NWMESSAGE Set enable_broker GO ALTER DATABASE NWDataChannel SET MULTI_USER GO */ SELECT 'Checking For Disabled Queues....' -- ensure the queues are enabled --  0 indicates the queue is disabled. Select '** Receive Queue Disabled: '+name from sys.service_queues where is_receive_enabled = 0 --select [name], is_receive_enabled from sys.service_queues; /*If the queue is disabled, to enable it alter queue QUEUENAME with status=on; – replace QUEUENAME with the name of your queue */ -- Get General information about the queues --select * from sys.service_queues -- Get the message counts in each queue SELECT 'Checking Message Count for each Queue...' select q.name, p.rows from sys.objects as o join sys.partitions as p on p.object_id = o.object_id join sys.objects as q on o.parent_object_id = q.object_id join sys.service_queues sq on sq.name = q.name where p.index_id = 1 -- Ensure all the queue activiation sprocs are present SELECT 'Checking for Activation Stored Procedures....' SELECT  '** Missing Procedure:  '+q.name  From sys.service_queues q Where NOT Exists(Select * from sysobjects where xtype='p' and name='activation_'+q.name) and q.activation_procedure is not null DECLARE @sprocs Table (Name Varchar(2000)) Insert into @sprocs Values ('Echo') Insert into @sprocs Values ('HTTP_POST') Insert into @sprocs Values ('InitializeRecipients') Insert into @sprocs Values ('sp_EnableRecipient') Insert into @sprocs Values ('sp_ProcessReceivedMessage') Insert into @sprocs Values ('sp_SendXmlMessage') SELECT 'Checking for required stored procedures...' SELECT  '** Missing Procedure:  '+s.name  From @sprocs s Where NOT Exists(Select * from sysobjects where xtype='p' and name=s.name) GO -- Check the services Select 'Checking Recipient Message Services...' Select '** Missing Message Service:' + r.RecipientName +'MessageService' From Recipient r Where not exists (Select * from sys.services s where  s.name  COLLATE SQL_Latin1_General_CP1_CI_AS= r.RecipientName+'MessageService') DECLARE @svcs Table (Name Varchar(2000)) Insert into @svcs Values ('XmlMessageSendingService') SELECT  '** Missing Service:  '+s.name  From @svcs s Where NOT Exists(Select * from sys.services where name=s.name COLLATE SQL_Latin1_General_CP1_CI_AS) GO /*** To Test a message send Run: sp_SendXmlMessage  'TSQLTEST', 'CommerceEngine','<Root><Text>Test</Text></Root>' */ Select CAST(message_body as XML) as xml, * From XmlMessageSendingQueue /*** clean out all queues declare @handle uniqueidentifier declare conv cursor for   select conversation_handle from sys.conversation_endpoints open conv fetch next from conv into @handle while @@FETCH_STATUS = 0 Begin    END Conversation @handle with cleanup    fetch next from conv into @handle End close conv deallocate conv ***********************

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  • view to select specific period or latest when null

    - by edosoft
    Hi I have a product table which simplifies to this: create table product(id int primary key identity, productid int, year int, quarter int, price money) and some sample data: insert into product select 11, 2010, 1, 1.11 insert into product select 11, 2010, 2, 2.11 insert into product select 11, 2010, 3, 3.11 insert into product select 12, 2010, 1, 1.12 insert into product select 12, 2010, 2, 2.12 insert into product select 13, 2010, 1, 1.13 Prices are can be changed each quarter, but not all products get a new price each quarter. Now I could duplicate the data each quarter, keeping the price the same, but I'd rather use a view. How can I create a view that can be used to return prices for (for example) quarter 2? I've written this to return the current (=latest) price: CREATE VIEW vwCurrentPrices AS SELECT * FROM ( SELECT *, ROW_NUMBER() OVER (PARTITION BY productid ORDER BY year DESC, quarter DESC) AS Ranking FROM product ) p WHERE p.Ranking = 1 I'd like to create a view so I can use queries like select * from vwProduct where quarter = 2

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  • jquery remove selected element and append to another

    - by KnockKnockWhosThere
    I'm trying to re-append a "removed option" to the appropriate select option menu. I have three select boxes: "Categories", "Variables", and "Target". "Categories" is a chained select, so when the user selects an option from it, the "Variables" select box is populated with options specific to the selected categories option. When the user chooses an option from the "Variables" select box, it's appended to the "Target" select box. I have a "remove selected" feature so that if a user "removes" a selected element from the "Target" select box, it's removed from "Target" and put back into the pool of "Variables" options. The problem I'm having is that it appends the option to the "Variables" items indiscriminately. That is, if the selected category is "Age" the "Variables" options all have a class of "age". But, if the removed option is an "income" item, it will display in the "Age Variables" option list. Here's the HTML markup: <select multiple="" id="categories" name="categories[]"> <option class="category" value="income">income</option> <option class="category" value="gender">gender</option> <option class="category" value="age">age</option> </select> <select multiple="multiple" id="variables" name="variables[]"> <option class="income" value="10">$90,000 - $99,999</option> <option class="income" value="11">$100,000 - $124,999</option> <option class="income" value="12">$125,000 - $149,999</option> <option class="income" value="13">Greater than $149,999</option> <option class="gender" value="14">Male</option> <option class="gender" value="15">Female</option> <option class="gender" value="16">Ungendered</option> <option class="age" value="17">Ages 18-24</option> <option class="age" value="18">Ages 25-34</option> <option class="age" value="19">Ages 35-44</option> </select> <select height="60" multiple="multiple" id="target" name="target[]"> </select> And, here's the js: /* This determines what options are display in the "Variables" select box */ var cat = $('#categories'); var el = $('#variables'); $('#categories option').click(function() { var class = $(this).val(); $('#variables option').each(function() { if($(this).hasClass(class)) { $(this).show(); } else { $(this).hide(); } }); }); /* This adds the option to the target select box if the user clicks "add" */ $('#add').click(function() { return !$('#variables option:selected').appendTo('#target'); }); /* This is the remove function in its current form, but doesn't append correctly */ $('#remove').click(function() { $('#target option:selected').each(function() { var class = $(this).attr('class'); if($('#variables option').hasClass(class)) { $(this).appendTo('#variables'); sortList('variables'); } }); });

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  • SQL SERVER – SELECT TOP Shortcut in SQL Server Management Studio (SSMS)

    - by pinaldave
    This is tool is pretty old, yet always comes as a handy tip. I had a great trip at TechEd in India. And, during one of my presentations, I was asked if there are any shortcuts to SELECT only TOP 100 records from SSMS. I immediately told him that if he explores the table in SSMS, he can just right click on it and SELECT TOP 1000 records. If he wanted only 100 records, then he could edit that 1000 to 100 by means of going to Options. Go to Options, then hover the mouse over the SQL Server Object Explorer, then proceed to Commands. Afterwards, change the Value for Select Top <n> Audit Records. After narrating the steps, he told me that he was not looking for the right click option; rather he was asking if there is any kind of keyboard shortcut for convenience’s sake. Actually, a keyboard shortcut is also possible. SQL Server Management Studio (SSMS) lets you configure the settings you want using a shortcut. Here is how you can do it. Go to Options, then to Environment. Proceed to Keyboard, and from there, configure your T-SQL with the desired keyword. Now, open SSMS New Query Window, and then click and type in any table name.  After that, just hit the shortcut you just made earlier. Doing this should display TOP 100 records in the Result window. I am sure this trick is quite old, but it is still helpful to many. Reference: Pinal Dave (http://blog.SQLAuthority.com) Filed under: Pinal Dave, SQL, SQL Add-On, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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