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  • Give back full control to a user on a disk from another computer

    - by Foghorn
    I have my friend's hard drive mounted externally. After messing with the permissions with TAKEOWN so I could fix some viruses, I have full control over their drive. The problem is, now it's stuck in a "autochk not found" reboot sequence. I think the problem is that the boot sector is invisible to the drive now. So my question is, How can I use icacls to give back the full ownership, when the user I am giving it to is not on my machine? I ran the TAKEOWN command from my windows 7 laptop, their machine is a windows xp Professional with three partitions, I only altered the one that has the boot sector. Here is the permissions that icacls shows: (Where my computer is %System% my username is ME, and the drive is E:\ C:\Users\ME icacls E:\* E:\$RECYCLE.BIN %System%\ME:(OI)(CI)(F) Mandatory Label\Low Mandatory Level:(OI)(CI)(IO)(NW) E:\ALLDATAW %System%\ME:(I)(OI)(CI)(F) E:\alrt_200.data %System%\ME:(OI)(CI)(F) E:\AUTOEXEC.BAT %System%\ME:(OI)(CI)(F) E:\AZ Commercial %System%\ME:(I)(OI)(CI)(F) E:\boot.ini %System%\ME:(OI)(CI)(F) E:\Config.Msi %System%\ME:(I)(OI)(CI)(F) E:\CONFIG.SYS %System%\ME:(OI)(CI)(F) E:\Documents and Settings %System%\ME:(I)(OI)(CI)(F) E:\IO.SYS %System%\ME:(OI)(CI)(F) E:\Mitchell1 %System%\ME:(I)(OI)(CI)(F) E:\MSDOS.SYS %System%\ME:(OI)(CI)(F) E:\MSOCache %System%\ME:(I)(OI)(CI)(F) E:\NTDClient.log %System%\ME:(OI)(CI)(F) E:\NTDETECT.COM %System%\ME:(OI)(CI)(F) E:\ntldr %System%\ME:(OI)(CI)(F) E:\pagefile.sys %System%\ME:(OI)(CI)(F) E:\Program Files %System%\ME:(I)(OI)(CI)(F) E:\RECYCLER %System%\ME:(I)(OI)(CI)(F) E:\RHDSetup.log %System%\ME:(OI)(CI)(F) E:\System Volume Information %System%\ME:(I)(OI)(CI)(F) E:\WINDOWS %System%\ME:(I)(OI)(CI)(F) Successfully processed 22 files; Failed processing 0 files C:\Users\ME

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  • Redundant Microsoft server solution for small company

    - by MadBoy
    I'm planning to change one server Microsoft SBS 2003 with SharePoint, Exchange and SQL database into something that will provide me with some redundancy and won't be single point of failure. I was thinking to buy 2x exactly the same physical servers and put 2 virtualized servers on HyperV or VMWare on each. Then i would put SharePoint, Exchange and SQL on that 1 physical server (shared onto 2x VM's). I would like 2nd physical server to be exact duplicate of the first one so that when 1st server goes down (for reboot or hw failure), 2nd takes care of everything so that users don't even see anything changed (in terms all their emails, sharepoint stuff is available). My questions are: Will I have to pay for licenses for both servers even thou only one instance of SharePoint, Exchange, SQL will be used at same time? What are proposed solutions to do that? Any additional hardware I would need, any complicated software configuration to be expected to configure such redundancy so that when one physical server goes down 2nd one is taking care of rest? What problems should I expect? This solution is for 60 people. Later on it may or may expand.

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  • MySQL Online Database

    - by Marian
    Can anyone suggest a good online free MySQL database. I've tried four till now: db4free FreeMySQL onPhP 000webhost Either of them gave me an timeout error on my connect file or actively restricted connection to it, meaning the host won't allow a remote connection to the database. If there isn't any good online database can I create my own server using my computer, since it gets rarely turned off and when my server is offline I could return an error message saying that the server is currently offline. My final objective is to have a simple comment box for a webpage. Witch I believe it won't need a massive data storage with 3 columns (id, name, comment) NOTE: Can't post more then two links yet.

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  • SQL SERVER Difference Between GRANT and WITH GRANT

    This was very interesting question recently asked me to during my session at TechMela Nepal. The question is what is the difference between GRANT and WITH GRANT when giving permissions to user.Let us first see syntax for the same.GRANT:USE master;GRANTVIEW ANY DATABASETO username;GOWITH GRANT:USE master;GRANTVIEW ANY DATABASETO username WITHGRANTOPTION;GOThe difference between both of this option [...]...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • The emergence of Atlassian's Bamboo (and a free SQL Source Control license offer!)

    - by David Atkinson
    The rise in demand for database continuous integration has forced me to skill-up in various new tools and technologies, particularly build servers. We have been using JetBrain's TeamCity here at Red Gate for a couple of years now, having replaced the ageing CruiseControl.NET, so it was a natural choice for us to use this for our database CI demos. Most of our early adopter customers have also transitioned away from CruiseControl, the majority to TeamCity and Microsoft's TeamBuild. However, more recently, for reasons we've yet to fully comprehend, we've observed a significant surge in the number of evaluators for Atlassian's Bamboo. I installed this a couple of weeks back to satisfy myself that it works seamlessly with Red Gate tools. As you would expect Bamboo's UI has the same clean feel found in any Atlassian tool (we use JIRA extensively here at Red Gate). In the coming weeks I will post a short step-by-step guide to setting up SQL Server continuous integration using the Red Gate command lines. To help us further optimize the integration between these tools I'd be very keen to hear from any Bamboo users who also use Red Gate tools who might be willing to participate in usability tests and other similar research in exchange for Amazon vouchers. If you are interested in helping out please contact me at David dot Atkinson at red-gate.com I recently spoke with Sarah, the product marketing manager for Bamboo, and we ended up having a detailed conversation about database CI, which has been meticulously documented in the form of a blog post on Atlassian's website: http://blogs.atlassian.com/2012/05/database-continuous-integration-redgate/ We've also managed to persuade Red Gate marketing to provide a great free-tool offer, provide a free SQL Source Control or SQL Connect license to Atlassian users provided it is claimed before the end of June! Full details are at the bottom of the post. Technorati Tags: sql server

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  • What are good replacements for Microsoft Visio 2003?

    - by James
    I have been using Visio 2003 on Windows XP for generation of UML diagrams. I have encountered following problems so far: There is no way to generate/print the documentation written for class attributes/methods. No automatic code generation is supported I have already generated lot of diagrams and i discovered above problems at much later stage. Now i would like to overcome above by choosing another tool which is compatible with Visio file(.vsd) which saves time or redrawing all diagrams and also provides above features. Could you kindly suggest an alternative (visio compatible) tool ? (I have looked at a similar SU-question, but it does not suggest tools which provide solution to above problems. I am open to free as well as licensed tools, with priority to free :) ) Thanks, James

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  • Daily tech links for .net and related technologies - June 8-11, 2010

    - by SanjeevAgarwal
    Daily tech links for .net and related technologies - June 8-11, 2010 Web Development ASPNET MVC: Handling Multiple Buttons on a Form with jQuery - Donn Building a MVC2 Template, Part 14, Logging Services - Eric Simple Accordion Menu With jQuery & ASP.NET - Steve Boschi Conditional Validation in MVC -Simonince Creating a RESTful Web Service Using ASP.Net MVC Part 23 – Bug Fixes and Area Support - Shoulders of Giants Web Design The Principles Of Cross-Browser CSS Coding - Louis Lazaris Transparency...(read more)

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  • The emergence of Atlassian's Bamboo (and a free SQL Source Control license offer!)

    - by David Atkinson
    The rise in demand for database continuous integration has forced me to skill-up in various new tools and technologies, particularly build servers. We have been using JetBrain's TeamCity here at Red Gate for a couple of years now, having replaced the ageing CruiseControl.NET, so it was a natural choice for us to use this for our database CI demos. Most of our early adopter customers have also transitioned away from CruiseControl, the majority to TeamCity and Microsoft's TeamBuild. However, more recently, for reasons we've yet to fully comprehend, we've observed a significant surge in the number of evaluators for Atlassian's Bamboo. I installed this a couple of weeks back to satisfy myself that it works seamlessly with Red Gate tools. As you would expect Bamboo's UI has the same clean feel found in any Atlassian tool (we use JIRA extensively here at Red Gate). In the coming weeks I will post a short step-by-step guide to setting up SQL Server continuous integration using the Red Gate command lines. To help us further optimize the integration between these tools I'd be very keen to hear from any Bamboo users who also use Red Gate tools who might be willing to participate in usability tests and other similar research in exchange for Amazon vouchers. If you are interested in helping out please contact me at David dot Atkinson at red-gate.com I recently spoke with Sarah, the product marketing manager for Bamboo, and we ended up having a detailed conversation about database CI, which has been meticulously documented in the form of a blog post on Atlassian's website: http://blogs.atlassian.com/2012/05/database-continuous-integration-redgate/ We've also managed to persuade Red Gate marketing to provide a great free-tool offer, provide a free SQL Source Control or SQL Connect license to Atlassian users provided it is claimed before the end of June! Full details are at the bottom of the post. Technorati Tags: sql server

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  • Swapping Function (Fn) and Control (Ctrl) Keys on Lenovo ThinkPad W500

    - by Howiecamp
    I'd like to swap the Fn and Ctrl keys on my ThinkPad W500 (like many others! See: How can I switch the function and control keys on my laptop? and Intercepting the Fn key on laptops) Numerous folks indicate that Windows doesn't register the Fn key as a keypress but using Mihov ASCII Master 2.0, that gives the ASCII value of a keypress, I see the Fn key returning FF (perhaps FF in this case means 'not registered'). I also see that keys like Ctrl register with one ASCII code when pressed alone and another when pressed in combo with another key. Fn will only register when pressed alone, so Windows definitely isn't seeing the combo. This took a solution like AutoHotKey off the table. I ran KeyTweak (which shows you the hardware scan codes of a keypress and the Fn key registerd as 57443). Using this program I remapped Fn to the Ctrl key; this worked perfectly. However, I suspect that because of the issue in #1, the combo of, for example, Fn + C did not execute a copy. Short of retraining my pinky I'm actually considering removing the keyboard and resoldering the connections to swap those keys. I'd love to get some input as to the root technical issue(s) and possible solutions here.

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  • CSS gradient not rendering in Windows Phone 8 WebBrowser Control

    - by SRSHawk
    I am facing an issue where, the CSS3 background is not rendered in WebBrowser control in Windows Phone 8. But same HTML when opened in WebBrowser in Windows Phone 8, it renders the gradient The HTML I am using is: <html> <head> <meta name="viewport" content="width=320, user-scalable=no, minimum-scale=1, maximum-scale=1"/> </head> <body style="margin:0px;overflow:hidden;"> <div id="im_c" style="height:48px;width:100%25; background: -ms-linear-gradient( bottom, #432100 30%, #00AAAA 70%);"> <div style="margin:0 auto;width:320px;"> Test </div> </div> <style> body {margin:0px} </style> </body> In Windows Phone 8, I use the HTML as below: WebBroswer WebView = new WebBrowser(); WebView.Height = 100; WebView.Width = 400; WebView.NavigateToString(@"<html><head><meta name=""viewport"" content=""width=320, user-scalable=no, minimum-scale=1, maximum-scale=1""/></head><body style=""margin:0px;overflow:hidden;""> <div id=""im_c"" style=""height:48px;width:100%25; background: -ms-linear-gradient( bottom, #432100 30%, #00AAAA 70%);""> <div style=""margin:0 auto;width:320px;"">Test</div></div> <style> body {margin:0px} </style> </body></html>"); In this case, the CSS gradient is not visible. Am I missing something?

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  • SQL SERVER Quick Note of Database Mirroring

    Just a day ago, I was invited at Round Table meeting at prestigious organization. They were planning to implement High Availability solution using Database Mirroring. During the meeting, I have made few notes of what was being discussed there. I just thought it would be interested for all of you know about it.Database Mirroring works [...]...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • Highly recommended: "5 Things SQL Server does different from what many developers expect" by Nico Ja

    A couple of weeks ago, the Belgian Techdays were held in Antwerp. Together with Scott Hillier I presented the SharePoint pre-conference sessions (watch them online over here, search for pre-conference or SharePoint). Even though Belgium is not a very big country, the Microsoft team managed to get some high profile speakers like Anders Hejlsberg and Scott Hanselman. But if you have like 60 minutes to spare there is one session that I'd really recommend to check out, not related to SharePoint, but...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • TSQL formatting - a sure fire way to start a conversation.

    - by fatherjack
    There are probably as many opinions on ways to format code as there are people writing code and I am not here to say that any one is better than any other. Well, that isn't true. I am here to say that one way is better than another but this isn't a matter of preference or personal taste, this is an example of where sloppy formatting can cause TSQL to weird and whacky things but following some simple methods can make your code more reliable and more robust when . Take these two pieces of code, ready...(read more)

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  • How to auto-install runlevel control for existing service/daemon?

    - by Johnny Utahh
    Need to install a service/daemon (in this case bind9, a DNS service) runlevel control, aka "rc" control (/etc/rc*.d and such). bind9 came pre-installed on my 11.04 system, but without aforementioned runlevel control. How to easily (and preferably automatically) install the rc stuff for "compliant" services/daemons in /etc/init.d? (Hint: I have the answer, but can't post it yet due to insufficient rep.)

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  • What is the difference between these two find algorithms? [migrated]

    - by Joe
    I have these two find algorithm which look the same to me. Can anyone help me out why they are actually different? Find ( x ) : if x.parent = x then return x else return Find ( x.parent ) vs Find ( x ) : if x.parent = x then return x else x.parent <- Find(x.parent) return x.parent I interpret the first one as int i = 0; return i++; while the second one as int i = 0; int tmp = i++; return tmp which are exactly the same to me.

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  • Basics of Join Factorization

    - by Hong Su
    We continue our series on optimizer transformations with a post that describes the Join Factorization transformation. The Join Factorization transformation was introduced in Oracle 11g Release 2 and applies to UNION ALL queries. Union all queries are commonly used in database applications, especially in data integration applications. In many scenarios the branches in a UNION All query share a common processing, i.e, refer to the same tables. In the current Oracle execution strategy, each branch of a UNION ALL query is evaluated independently, which leads to repetitive processing, including data access and join. The join factorization transformation offers an opportunity to share the common computations across the UNION ALL branches. Currently, join factorization only factorizes common references to base tables only, i.e, not views. Consider a simple example of query Q1. Q1:    select t1.c1, t2.c2    from t1, t2, t3    where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c2 = 2 and t2.c2 = t3.c2   union all    select t1.c1, t2.c2    from t1, t2, t4    where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c3 = t4.c3; Table t1 appears in both the branches. As does the filter predicates on t1 (t1.c1 > 1) and the join predicates involving t1 (t1.c1 = t2.c1). Nevertheless, without any transformation, the scan (and the filtering) on t1 has to be done twice, once per branch. Such a query may benefit from join factorization which can transform Q1 into Q2 as follows: Q2:    select t1.c1, VW_JF_1.item_2    from t1, (select t2.c1 item_1, t2.c2 item_2                   from t2, t3                    where t2.c2 = t3.c2 and t2.c2 = 2                                  union all                   select t2.c1 item_1, t2.c2 item_2                   from t2, t4                    where t2.c3 = t4.c3) VW_JF_1    where t1.c1 = VW_JF_1.item_1 and t1.c1 > 1; In Q2, t1 is "factorized" and thus the table scan and the filtering on t1 is done only once (it's shared). If t1 is large, then avoiding one extra scan of t1 can lead to a huge performance improvement. Another benefit of join factorization is that it can open up more join orders. Let's look at query Q3. Q3:    select *    from t5, (select t1.c1, t2.c2                  from t1, t2, t3                  where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c2 = 2 and t2.c2 = t3.c2                 union all                  select t1.c1, t2.c2                  from t1, t2, t4                  where t1.c1 = t2.c1 and t1.c1 > 1 and t2.c3 = t4.c3) V;   where t5.c1 = V.c1 In Q3, view V is same as Q1. Before join factorization, t1, t2 and t3 must be joined first before they can be joined with t5. But if join factorization factorizes t1 from view V, t1 can then be joined with t5. This opens up new join orders. That being said, join factorization imposes certain join orders. For example, in Q2, t2 and t3 appear in the first branch of the UNION ALL query in view VW_JF_1. T2 must be joined with t3 before it can be joined with t1 which is outside of the VW_JF_1 view. The imposed join order may not necessarily be the best join order. For this reason, join factorization is performed under cost-based transformation framework; this means that we cost the plans with and without join factorization and choose the cheapest plan. Note that if the branches in UNION ALL have DISTINCT clauses, join factorization is not valid. For example, Q4 is NOT semantically equivalent to Q5.   Q4:     select distinct t1.*      from t1, t2      where t1.c1 = t2.c1  union all      select distinct t1.*      from t1, t2      where t1.c1 = t2.c1 Q5:    select distinct t1.*     from t1, (select t2.c1 item_1                   from t2                union all                   select t2.c1 item_1                  from t2) VW_JF_1     where t1.c1 = VW_JF_1.item_1 Q4 might return more rows than Q5. Q5's results are guaranteed to be duplicate free because of the DISTINCT key word at the top level while Q4's results might contain duplicates.   The examples given so far involve inner joins only. Join factorization is also supported in outer join, anti join and semi join. But only the right tables of outer join, anti join and semi joins can be factorized. It is not semantically correct to factorize the left table of outer join, anti join or semi join. For example, Q6 is NOT semantically equivalent to Q7. Q6:     select t1.c1, t2.c2    from t1, t2    where t1.c1 = t2.c1(+) and t2.c2 (+) = 2  union all    select t1.c1, t2.c2    from t1, t2      where t1.c1 = t2.c1(+) and t2.c2 (+) = 3 Q7:     select t1.c1, VW_JF_1.item_2    from t1, (select t2.c1 item_1, t2.c2 item_2                  from t2                  where t2.c2 = 2                union all                  select t2.c1 item_1, t2.c2 item_2                  from t2                                                                                                    where t2.c2 = 3) VW_JF_1       where t1.c1 = VW_JF_1.item_1(+)                                                                  However, the right side of an outer join can be factorized. For example, join factorization can transform Q8 to Q9 by factorizing t2, which is the right table of an outer join. Q8:    select t1.c2, t2.c2    from t1, t2      where t1.c1 = t2.c1 (+) and t1.c1 = 1 union all    select t1.c2, t2.c2    from t1, t2    where t1.c1 = t2.c1(+) and t1.c1 = 2 Q9:   select VW_JF_1.item_2, t2.c2   from t2,             (select t1.c1 item_1, t1.c2 item_2            from t1            where t1.c1 = 1           union all            select t1.c1 item_1, t1.c2 item_2            from t1            where t1.c1 = 2) VW_JF_1   where VW_JF_1.item_1 = t2.c1(+) All of the examples in this blog show factorizing a single table from two branches. This is just for ease of illustration. Join factorization can factorize multiple tables and from more than two UNION ALL branches.  SummaryJoin factorization is a cost-based transformation. It can factorize common computations from branches in a UNION ALL query which can lead to huge performance improvement. 

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  • Serial plans: Threshold / Parallel_degree_limit = 1

    - by jean-pierre.dijcks
    As a very short follow up on the previous post. So here is some more on getting a serial plan and why that happens Another reason - compared to the auto DOP is not on as we looked at in the earlier post - and often more prevalent to get a serial plan is if the plan simply does not take long enough to consider a parallel path. The resulting plan and note looks like this (note that this is a serial plan!): explain plan for select count(1) from sales; SELECT PLAN_TABLE_OUTPUT FROM TABLE(DBMS_XPLAN.DISPLAY()); PLAN_TABLE_OUTPUT -------------------------------------------------------------------------------- Plan hash value: 672559287 -------------------------------------------------------------------------------------- | Id  | Operation            | Name  | Rows  | Cost (%CPU)| Time     | Pstart| Pstop | -------------------------------------------------------------------------------------- PLAN_TABLE_OUTPUT -------------------------------------------------------------------------------- |   0 | SELECT STATEMENT     |       |     1 |     5   (0)| 00:00:01 |       |     | |   1 |  SORT AGGREGATE      |       |     1 |            |          |       |     | |   2 |   PARTITION RANGE ALL|       |   960 |     5   (0)| 00:00:01 |     1 |  16 | |   3 |    TABLE ACCESS FULL | SALES |   960 |     5   (0)| 00:00:01 |     1 |  16 | Note -----    - automatic DOP: Computed Degree of Parallelism is 1 because of parallel threshold 14 rows selected. The parallel threshold is referring to parallel_min_time_threshold and since I did not change the default (10s) the plan is not being considered for a parallel degree computation and is therefore staying with the serial execution. Now we go into the land of crazy: Assume I do want this DOP=1 to happen, I could set the parameter in the init.ora, but to highlight it in this case I changed it on the session: alter session set parallel_degree_limit = 1; The result I get is: ERROR: ORA-02097: parameter cannot be modified because specified value is invalid ORA-00096: invalid value 1 for parameter parallel_degree_limit, must be from among CPU IO AUTO INTEGER>=2 Which of course makes perfect sense...

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  • Proper Data Structure for Commentable Comments

    - by Wesley
    Been struggling with this on an architectural level. I have an object which can be commented on, let's call it a Post. Every post has a unique ID. Now I want to comment on that Post, and I can use ID as a foreign key, and each PostComment has an ItemID field which correlates to the Post. Since each Post has a unique ID, it is very easy to assign "Top Level" comments. When I comment on a comment however, I feel like I now need a PostCommentComment, which attaches to the ID of the PostComment. Since ID's are assigned sequentially, I can no longer simply use ItemID to differentiate where in the tree the comment is assigned. I.E. both a Post and a Post Comment might have an ID of '5', so my foreign key relationship is invalid. This seems like it could go on infinitely, with PostCommentCommentComment's etc... What's the best way to solve this? Should I have a field in the comment called "IsPostComment" or something of the like to know which collection to attach the ID to? This strikes me as the best solution I've seen so far, but now I feel like I need to make recursive DataBase calls which start to get expensive. Meaning, I get a Post and get all PostComments where ItemID == Post.ID && where IsPostComment == true Then I take that as a collection, gather all the ID's of the PostComments, and do another search where ItemID == PostComment[all].ID && where IsPostComment == false, then repeat infinitely. This means I make a call for every layer, and if I'm calling 100 Posts, I might make 1000 DB calls to get 10 layers of comments each. What is the right way to do this?

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  • Swapping Function (Fn) and Control (Ctrl) Keys on Lenovo ThinkPad W500

    - by Howiecamp
    I'd like to swap the Fn and Ctrl keys on my ThinkPad W500 (like many others!). I wanted to comment on http://superuser.com/questions/35228/how-can-i-switch-the-function-and-control-keys-on-my-laptop and StackOverflow question 514781 (please Google it because I don't have enough rep to include 2 hyperlinks) but I don't have enough rep to do so to comment. Numerous folks (in both the above questions and on other Google searches) indicate that Windows doesn't register the Fn key as a keypress but using a tool that gives the ASCII value of a keypress (visit www mihov com / eng / am.html) I see the Fn key returning FF (perhaps FF in this case means 'not registered'). I also see that keys like Ctrl register with one ASCII code when pressed alone and another when pressed in combo with another key. Fn will only register when pressed alone, so Windows definitely isn't seeing the combo. This took a solution like AutoHotKey off the table. I ran KeyTweak (which shows you the hardware scan codes of a keypress and the Fn key registerd as 57443). Using this program I remapped Fn to the Ctrl key; this worked perfectly. However, I suspect that because of the issue in #1, the combo of, for example, Fn + C did not execute a copy. Short of retraining my pinky I'm actually considering removing the keyboard and resoldering the connections to swap those keys. I'd love to get some input as to the root technical issue(s) and possible solutions here. Thanks

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  • MSSQLSERVER Will Not Start - Event ID 913 and 1814

    - by ThaKidd
    Hello ServerFault! I need some serious help. I have a major database server down and am scratching my head at how to fix it. The server was hit by rolling black outs last week in Dallas and sense then, Microsoft SQL 2005 SP2 will not start up. I am getting the following errors (both when starting the service and while trying to execute mssqlsrv.exe -c -f -m: Event Type: Error Event Source: MSSQLSERVER Event ID: 913 Could not find database ID 3. Database may not be activated yet or may be in transition. Reissue the query once the database is available. If you do not think this error is due to a database that is transitioning its state and this error continues to occur, contact your primary support provider. Please have available for review the Microsoft SQL Server error log and any additional information relevant to the circumstances when the error occurred. and... Event Type: Information Event Source: MSSQLSERVER Event ID: 1814 Could not create tempdb. You may not have enough disk space available. Free additional disk space by deleting other files on the tempdb drive and then restart SQL Server. Check for additional errors in the event log that may indicate why the tempdb files could not be initialized. I have tried to rename the tempdb.mdf to tempdb.old with no success. I have checked and have 193 GB of free hard drive space. What else might cause this problem? Could the server need a chkdsk ran on it or do I need to be looking at some area of the database server? Any help is greatly appreciated. Thank you in advance.

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  • Materialized View does not import properly when importing a db for a second time in another schema [closed]

    - by marinus
    When I import a database with materialized view mv_mt in just one schema there are no errors. create materialized view mv_mt refresh complete next trunc( sysdate ) + 1 as SELECT sysdate, media_type.* from media_type; But when I try to import the same database to a copy in another schema in the same tablespace I get the following errors: IMP-00017: following statement failed with ORACLE error 1: "BEGIN DBMS_JOB.ISUBMIT(JOB=>438,WHAT=>'dbms_refresh.refresh(''"ALEXANDRA"" "."MV_MT"'');',NEXT_DATE=>TO_DATE('2012-07-02:14:22:36','YYYY-MM-DD:HH24:MI:" "SS'),INTERVAL=>'sysdate + 1 / 24 / 60 / 6 ',NO_PARSE=>TRUE); END;" IMP-00003: ORACLE error 1 encountered ORA-00001: unique constraint (SYS.I_JOB_JOB) violated ORA-06512: at "SYS.DBMS_JOB", line 100 ORA-06512: at line 1 IMP-00017: following statement failed with ORACLE error 23421: "BEGIN dbms_refresh.make('"ALEXANDRA"."MV_MT"',list=>null,next_date=>null," "interval=>null,implicit_destroy=>TRUE,lax=>FALSE,job=>438,rollback_seg=>NUL" "L,push_deferred_rpc=>TRUE,refresh_after_errors=>FALSE,purge_option => 1,par" "allelism => 0,heap_size => 0); END;" IMP-00003: ORACLE error 23421 encountered ORA-23421: job number 438 is not a job in the job queue ORA-06512: at "SYS.DBMS_SYS_ERROR", line 86 ORA-06512: at "SYS.DBMS_IJOB", line 793 ORA-06512: at "SYS.DBMS_REFRESH", line 86 ORA-06512: at "SYS.DBMS_REFRESH", line 62 ORA-06512: at line 1 IMP-00017: following statement failed with ORACLE error 23410: "BEGIN dbms_refresh.add(name=>'"ALEXANDRA"."MV_MT"',list=>'"ALEXANDRA"."MV" "_MT"',siteid=>0,export_db=>'ORCL01'); END;" IMP-00003: ORACLE error 23410 encountered ORA-23410: materialized view "ALEXANDRA"."MV_MT" is already in a refresh group ORA-06512: at "SYS.DBMS_SYS_ERROR", line 95 ORA-06512: at "SYS.DBMS_IREFRESH", line 484 ORA-06512: at "SYS.DBMS_REFRESH", line 140 ORA-06512: at "SYS.DBMS_REFRESH", line 125 ORA-06512: at line 1 Anyone any ideas?

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  • Adding Actions to a Cube in SQL Server Analysis Services 2008

    Actions are powerful way of extending the value of SSAS cubes for the end user. They can click on a cube or portion of a cube to start an application with the selected item as a parameter, or to retrieve information about the selected item. Actions haven't been well-documented until now; Robert Sheldon once more makes everything clear.

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