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  • SQL SERVER – Find Referenced or Referencing Object in SQL Server using sys.sql_expression_dependencies

    - by pinaldave
    A very common question which I often receive are: How do I find all the tables used in a particular stored procedure? How do I know which stored procedures are using a particular table? Both are valid question but before we see the answer of this question – let us understand two small concepts – Referenced and Referencing. Here is the sample stored procedure. CREATE PROCEDURE mySP AS SELECT * FROM Sales.Customer GO Reference: The table Sales.Customer is the reference object as it is being referenced in the stored procedure mySP. Referencing: The stored procedure mySP is the referencing object as it is referencing Sales.Customer table. Now we know what is referencing and referenced object. Let us run following queries. I am using AdventureWorks2012 as a sample database. If you do not have SQL Server 2012 here is the way to get SQL Server 2012 AdventureWorks database. Find Referecing Objects of a particular object Here we are finding all the objects which are using table Customer in their object definitions (regardless of the schema). USE AdventureWorks GO SELECT referencing_schema_name = SCHEMA_NAME(o.SCHEMA_ID), referencing_object_name = o.name, referencing_object_type_desc = o.type_desc, referenced_schema_name, referenced_object_name = referenced_entity_name, referenced_object_type_desc = o1.type_desc, referenced_server_name, referenced_database_name --,sed.* -- Uncomment for all the columns FROM sys.sql_expression_dependencies sed INNER JOIN sys.objects o ON sed.referencing_id = o.[object_id] LEFT OUTER JOIN sys.objects o1 ON sed.referenced_id = o1.[object_id] WHERE referenced_entity_name = 'Customer' The above query will return all the objects which are referencing the table Customer. Find Referenced Objects of a particular object Here we are finding all the objects which are used in the view table vIndividualCustomer. USE AdventureWorks GO SELECT referencing_schema_name = SCHEMA_NAME(o.SCHEMA_ID), referencing_object_name = o.name, referencing_object_type_desc = o.type_desc, referenced_schema_name, referenced_object_name = referenced_entity_name, referenced_object_type_desc = o1.type_desc, referenced_server_name, referenced_database_name --,sed.* -- Uncomment for all the columns FROM sys.sql_expression_dependencies sed INNER JOIN sys.objects o ON sed.referencing_id = o.[object_id] LEFT OUTER JOIN sys.objects o1 ON sed.referenced_id = o1.[object_id] WHERE o.name = 'vIndividualCustomer' The above query will return all the objects which are referencing the table Customer. I am just glad to write above query. There are more to write to this subject. In future blog post I will write more in depth about other DMV which also aids in finding referenced data. Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: PostADay, SQL, SQL Authority, SQL DMV, SQL Query, SQL Server, SQL Tips and Tricks, SQL Utility, T SQL, Technology

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  • Am 10.02. startet WebCast-Serie für Java Entwickler und WebLogic Interessenten: WebLogic Developer - Get the latest on Oracle WebLogic Server and Java EE 6

    - by Thomas Leopold
    v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} Normal 0 21 false false false DE X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0cm 5.4pt 0cm 5.4pt; mso-para-margin:0cm; mso-para-margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi;} Accelerate Your Development with Oracle WebLogic Suite Many organisations are reducing travel, conference, and training budgets for their developers without any change to the results expected of those developers. So how can you keep up with the latest developments?By receiving training, delivered free of charge, at your desk!Join us during February and March for a series of online events designed and run by the development team at Oracle. Learn how Oracle WebLogic Suite enables a whole new level of productivity for enterprise developers.Virtual Developer Day - 10th FebruaryStarting with our Virtual Developer Day on 10th February, join us for a blend of hands-on labs, live chat and presentations covering the latest on WebLogic, Java EE 6 and the programming tenets that have made it a true platform breakthrough.Weekly WebLogic Webcasts from 17th February to 17th MarchAfterwards, join us every week from 17th February to 17th March for our weekly one-hour webcasts where we will show you how to build an application from the ground up using Java and JEE technologies. Presented by the engineering team for WebLogic, these webcasts will be of great value to developers and architects, not just those already using WebLogic.For registration, full session abstracts and schedule please click here. Don't miss out! Register now to join our virtual events and keep up with all the latest developments. Find out more and register now Copyright © 2011, Oracle Corporation and/or its affiliates.All rights reserved. Contact Us | Legal Notices and Terms of Use | Privacy Statement

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  • Likewise: joined Active Directory but cannot write shares.

    - by Aron Rotteveel
    I have never used a Linux system in an AD environment before and am trying to join my laptop running Ubuntu to join our Active Directory (DC is a Windows Server 2008 machine) using Likewise-open. Using the GUI wizard, I have joined the domain. I can mount network shares using CIFS Problem: I only have read access to our fileserver. What more is needed to get the AD to recognize me as a user who has the appropriate rights? Any help is appreciated.

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  • Google.org Crisis Response and the Google Maps APIs

    Google.org Crisis Response and the Google Maps APIs This week, Pete Giencke and Ka-Ping Yee of the Google Crisis Response Team join Paul Saxman to talk about the technologies and data they use for their mapping efforts, such as the Crisis Map and Google Public Alerts. Join us to learn how to use the Google Maps APIs to track hurricanes, monitor floods, and help affected users locate critical information such as shelters and evacuation routes in the aftermath of a disaster. From: GoogleDevelopers Views: 0 0 ratings Time: 00:00 More in Science & Technology

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  • Free Virtual Developer Day: Oracle Fusion Development on July, 10th

    - by Lionel Dubreuil
    Simpler Java Development with Oracle ADF and Fusion Middleware. Join a free online developer day where you can learn about the various components that make up the Oracle Fusion Middleware development platform including Oracle WebCenter, Business Intelligence, BPM and more! Online seminars, hands-on lab and live chats with our technical staff is available directly from your computer.  Register now and join us on July 10th: https://oracle.6connex.com/portal/fusiondev/login?langR=en_US

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  • Free Virtual Developer Day: Oracle Fusion Development on July, 10th

    - by Lionel Dubreuil
    Simpler Java Development with Oracle ADF and Fusion Middleware. Join a free online developer day where you can learn about the various components that make up the Oracle Fusion Middleware development platform including Oracle WebCenter, Business Intelligence, BPM and more! Online seminars, hands-on lab and live chats with our technical staff is available directly from your computer.  Register now and join us on July 10th: https://oracle.6connex.com/portal/fusiondev/login?langR=en_US

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  • Happy Birthday, SQLPeople!

    - by andyleonard
    One year ago today, I began sending out batches of SQLPeople interview emails to friends in the SQL Server Community. Since then, Brian Moran ( Blog | @briancmoran ) and Matt Velic ( Blog | @mvelic | SQLPeople ) have joined the effort, we have published dozens of interviews, and there have been two events! You can join in the fun. If you haven’t already, visit the interview page and answer the seven questions. You can also join us on LinkedIn and Facebook . And you can follow us on Twitter ( @SQLPeople...(read more)

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  • Free Webinar - Using Enterprise Data Integration Dashboards

    - by andyleonard
    Join Kent Bradshaw and me as we present Using Enterprise Data Integration Dashboards Tuesday 11 Dec 2012 at 10:00 AM ET! If data is the life of the modern organization, data integration is the heart of an enterprise. Data circulation is vital. Data integration dashboards provide enterprise ETL (Extract, Transform, and Load) teams near-real-time status supported with historical performance analysis. Join Linchpins Kent Bradshaw and Andy Leonard as they demonstrate and discuss the benefits of data...(read more)

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  • Stairway to T-SQL DML Level 5: The Mathematics of SQL: Part 2

    Joining tables is a crucial concept to understanding data relationships in a relational database. When you are working with your SQL Server data, you will often need to join tables to produce the results your application requires. Having a good understanding of set theory, and the mathematical operators available and how they are used to join tables will make it easier for you to retrieve the data you need from SQL Server.

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  • Spring SQL Connections 2011 and SQLServerCentral.

    Once again SQLServerCentral is sponsoring a track at SQL Connections in Orlando this March. Read about the event and our speakers and join us for SQL Server training in Florida. Join SQL Backup’s 35,000+ customers to compress and strengthen your backups "SQL Backup will be a REAL boost to any DBA lucky enough to use it." Jonathan Allen. Download a free trial now.

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  • problem with sIFR 3 not displaying in IE just getting XXX

    - by user288306
    I am having a problem with sIFR 3 not displaying in IE. I get 3 larges black XXX in IE yet it displays fine in Firefox. I have checked i do have the most recent version of flash installed correctly. Here is the code on the page <div id="features"> <div id="mainmessage_advertisers"><h2>Advertisers</h2><br /><br /><h3><a href="">Reach your customers where they browse. Buy directly from top web publishers.</a></h3><br /><br /><br /><a href=""><img src="img/buyads.gif" border="0"></a></div> <div id="mainmessage_publishers"><h2>Publishers</h2><br /><br /><h3>Take control of your ad space and start generating more revenue than <u>ever before</u>.</h3><br /><br /><br /><a href=""><img src="img/sellads.gif" border="0"></a></div> </div>` Here is the code from my global.css #mainmessage_advertisers { width: 395px; height: 200px; padding: 90px 50px; border: 1px; float: left; } #mainmessage_publishers { width: 395px; height: 200px; padding: 90px 50px; float: right; } and here is what i have in my sifr.js /*********************************************************************** SIFR 3.0 (BETA 1) FUNCTIONS ************************************************************************/ var parseSelector=(function(){var _1=/\s*,\s*/;var _2=/\s*([\s>+~(),]|^|$)\s*/g;var _3=/([\s>+~,]|[^(]\+|^)([#.:@])/g;var _4=/^[^\s>+~]/;var _5=/[\s#.:>+~()@]|[^\s#.:>+~()@]+/g;function 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  • Create SAMBA node trust relationship to Windows 2003 PDC server

    - by Rod Regier
    I am having problems creating a trust relationship between an OpenVMS/IA64 node running V/IA64 8.3-1H1, TCPIP 5.6 ECO 5, CIFS 1.1 ECO1 PS11 (SAMBA 3.0.28a) and Windows 2003 server running as a PDC. I do have two other OpenVMS/Alpha nodes running V/A 8.3, TCPIP 5.6 ECO 4, CIS 1.1 ECO1 PS10 (SAMBA 3.0.28a) with working trust relationships to the same Windows 2003 server. Looking for assistance in resolving the trust "handshake". \\ Details from failing node. Unless otherwise noted, corresponding files on working nodes are similar or identical. SMB.CONF extract: [global] server string = Samba %v running on %h (OpenVMS) workgroup = WILMA netbios name = %h security = DOMAIN encrypt passwords = Yes name resolve order = lmhosts host wins bcast Password server = * log file = /samba$log/log.%m printcap name = /sys$manager/ucx$printcap.dat guest account = DYMAX print command = print %f/queue=%p/delete/passall/name="""""%s""""" lprm command = delete/entry=%j map archive = No printing = OpenVMS net rpc testjoin [2010/08/13 16:09:28, 0] SAMBA$SRC:[SOURCE.RPC_CLIENT]CLI_PIPE.C;1:(2443) get_schannel_session_key: could not fetch trust account password for domain 'WILMA' [2010/08/13 16:09:28, 0] SAMBA$SRC:[SOURCE.UTILS]NET_RPC_JOIN.C;1:(72) net_rpc_join_ok: failed to get schannel session key from server W2K3AD2 for domain WILMA. Error was NT_STATUS_CANT_ACCESS_DOMAIN_I NFO Join to domain 'WILMA' is not valid net rpc join "-Uaccount%password" tdb_open_isam: error verifying status of file SAMBA$ROOT:[PRIVATE]secrets.tdb tdb_open_isam: errno value = 1 [2010/08/13 16:21:13, 0] SAMBA$SRC:[SOURCE.PASSDB]SECRETS.C;1:(72) Failed to open /SAMBA$ROOT/PRIVATE/secrets.tdb [2010/08/13 16:21:13, 0] SAMBA$SRC:[SOURCE.UTILS]NET_RPC.C;1:(322) error storing domain sid for WILMA tdb_open_isam: error verifying status of file SAMBA$ROOT:[PRIVATE]secrets.tdb tdb_open_isam: errno value = 1 [2010/08/13 16:21:13, 0] SAMBA$SRC:[SOURCE.PASSDB]SECRETS.C;1:(72) Failed to open /SAMBA$ROOT/PRIVATE/secrets.tdb [2010/08/13 16:21:13, 0] SAMBA$SRC:[SOURCE.UTILS]NET_RPC_JOIN.C;1:(409) error storing domain sid for WILMA Unable to join domain WILMA. \\ Example from other node: net rpc testjoin Join to 'WILMA' is OK

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  • Slow INFORMATION_SCHEMA query

    - by Thomas
    We have a .NET Windows application that runs the following query on login to get some information about the database: SELECT t.TABLE_NAME, ISNULL(pk_ccu.COLUMN_NAME,'') PK, ISNULL(fk_ccu.COLUMN_NAME,'') FK FROM INFORMATION_SCHEMA.TABLES t LEFT JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS pk_tc ON pk_tc.TABLE_NAME = t.TABLE_NAME AND pk_tc.CONSTRAINT_TYPE = 'PRIMARY KEY' LEFT JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE pk_ccu ON pk_ccu.CONSTRAINT_NAME = pk_tc.CONSTRAINT_NAME LEFT JOIN INFORMATION_SCHEMA.TABLE_CONSTRAINTS fk_tc ON fk_tc.TABLE_NAME = t.TABLE_NAME AND fk_tc.CONSTRAINT_TYPE = 'FOREIGN KEY' LEFT JOIN INFORMATION_SCHEMA.CONSTRAINT_COLUMN_USAGE fk_ccu ON fk_ccu.CONSTRAINT_NAME = fk_tc.CONSTRAINT_NAME Usually this runs in a couple seconds, but on one server running SQL Server 2000, it is taking over four minutes to run. I ran it with the execution plan enabled, and the results are huge, but this part caught my eye (it won't let me post an image): http://img35.imageshack.us/i/plank.png/ I then updated the statistics on all of the tables that were mentioned in the execution plan: update statistics sysobjects update statistics syscolumns update statistics systypes update statistics master..spt_values update statistics sysreferences But that didn't help. The index tuning wizard doesn't help either, because it doesn't let me select system tables. There is nothing else running on this server, so nothing else could be slowing it down. What else can I do to diagnose or fix the problem on that server?

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  • Domain joining debate for Outlook 2010 with Exchange 2007 on windows SBS 2008 for a user on a laptop that will travel a fair amount of the time.

    - by user71195
    I'm basically debating on whether or not to join the Domain on a Laptop, and was wondering if anyone has had a similar experience. If the computer were staying in the office, its a no brainer. Join the domain. In this case I have a user who will come into the office a few days a week, and work remotely the rest of the time. There is a working VPN using OpenVPN client/server, but it's not site-to-site. My knee jerk reaction is to not join the domain, so that the user can have 1 profile that they always use. In this configuration, should Outlook work properly with the user's domain account, and should the shared calendar still work (at least once inside the VPN)? My concern with joining the domain would be the inability to login to it when elsewhere. Is there maybe a way around this with caching or something? Would creating a second local login make sense for a user like this in any way? If so, why not just skip the domain join to begin with? Any thoughts on or experiences with this would be appreciated. Laptop OS Windows 7 (Not purchased yet.. pro if domain needed) Server SBS 2008, Exchange 2007 Outlook version 2010 Thanks for any help, Mike

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  • Nesting Linq-to-Objects query within Linq-to-Entities query –what is happening under the covers?

    - by carewithl
    var numbers = new int[] { 1, 2, 3, 4, 5 }; var contacts = from c in context.Contacts where c.ContactID == numbers.Max() | c.ContactID == numbers.FirstOrDefault() select c; foreach (var item in contacts) Console.WriteLine(item.ContactID); Linq-to-Entities query is first translated into Linq expression tree, which is then converted by Object Services into command tree. And if Linq-to-Entities query nests Linq-to-Objects query, then this nested query also gets translated into an expression tree. a) I assume none of the operators of the nested Linq-to-Objects query actually get executed, but instead data provider for particular DB (or perhaps Object Services) knows how to transform the logic of Linq-to-Objects operators into appropriate SQL statements? b) Data provider knows how to create equivalent SQL statements only for some of the Linq-to-Objects operators? c) Similarly, data provider knows how to create equivalent SQL statements only for some of the non-Linq methods in the Net Framework class library? EDIT: I know only some Sql so I can't be completely sure, but reading Sql query generated for the above code it seems data provider didn't actually execute numbers.Max method, but instead just somehow figured out that numbers.Max should return the maximum value and then proceed to include in generated Sql query a call to TSQL's build-in MAX function. It also put all the values held by numbers array into a Sql query. SELECT CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN '0X0X' ELSE '0X1X' END AS [C1], [Extent1].[ContactID] AS [ContactID], [Extent1].[FirstName] AS [FirstName], [Extent1].[LastName] AS [LastName], [Extent1].[Title] AS [Title], [Extent1].[AddDate] AS [AddDate], [Extent1].[ModifiedDate] AS [ModifiedDate], [Extent1].[RowVersion] AS [RowVersion], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[CustomerTypeID] END AS [C2], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[InitialDate] END AS [C3], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[PrimaryDesintation] END AS [C4], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[SecondaryDestination] END AS [C5], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[PrimaryActivity] END AS [C6], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[SecondaryActivity] END AS [C7], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[Notes] END AS [C8], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[RowVersion] END AS [C9], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[BirthDate] END AS [C10], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[HeightInches] END AS [C11], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[WeightPounds] END AS [C12], CASE WHEN (([Project1].[C1] = 1) AND ([Project1].[C1] IS NOT NULL)) THEN [Project1].[DietaryRestrictions] END AS [C13] FROM [dbo].[Contact] AS [Extent1] LEFT OUTER JOIN (SELECT [Extent2].[ContactID] AS [ContactID], [Extent2].[BirthDate] AS [BirthDate], [Extent2].[HeightInches] AS [HeightInches], [Extent2].[WeightPounds] AS [WeightPounds], [Extent2].[DietaryRestrictions] AS [DietaryRestrictions], [Extent3].[CustomerTypeID] AS [CustomerTypeID], [Extent3].[InitialDate] AS [InitialDate], [Extent3].[PrimaryDesintation] AS [PrimaryDesintation], [Extent3].[SecondaryDestination] AS [SecondaryDestination], [Extent3].[PrimaryActivity] AS [PrimaryActivity], [Extent3].[SecondaryActivity] AS [SecondaryActivity], [Extent3].[Notes] AS [Notes], [Extent3].[RowVersion] AS [RowVersion], cast(1 as bit) AS [C1] FROM [dbo].[ContactPersonalInfo] AS [Extent2] INNER JOIN [dbo].[Customers] AS [Extent3] ON [Extent2].[ContactID] = [Extent3].[ContactID]) AS [Project1] ON [Extent1].[ContactID] = [Project1].[ContactID] LEFT OUTER JOIN (SELECT TOP (1) [c].[C1] AS [C1] FROM (SELECT [UnionAll3].[C1] AS [C1] FROM (SELECT [UnionAll2].[C1] AS [C1] FROM (SELECT [UnionAll1].[C1] AS [C1] FROM (SELECT 1 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable1] UNION ALL SELECT 2 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable2]) AS [UnionAll1] UNION ALL SELECT 3 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable3]) AS [UnionAll2] UNION ALL SELECT 4 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable4]) AS [UnionAll3] UNION ALL SELECT 5 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable5]) AS [c]) AS [Limit1] ON 1 = 1 LEFT OUTER JOIN (SELECT TOP (1) [c].[C1] AS [C1] FROM (SELECT [UnionAll7].[C1] AS [C1] FROM (SELECT [UnionAll6].[C1] AS [C1] FROM (SELECT [UnionAll5].[C1] AS [C1] FROM (SELECT 1 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable6] UNION ALL SELECT 2 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable7]) AS [UnionAll5] UNION ALL SELECT 3 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable8]) AS [UnionAll6] UNION ALL SELECT 4 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable9]) AS [UnionAll7] UNION ALL SELECT 5 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable10]) AS [c]) AS [Limit2] ON 1 = 1 CROSS JOIN (SELECT MAX([UnionAll12].[C1]) AS [A1] FROM (SELECT [UnionAll11].[C1] AS [C1] FROM (SELECT [UnionAll10].[C1] AS [C1] FROM (SELECT [UnionAll9].[C1] AS [C1] FROM (SELECT 1 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable11] UNION ALL SELECT 2 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable12]) AS [UnionAll9] UNION ALL SELECT 3 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable13]) AS [UnionAll10] UNION ALL SELECT 4 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable14]) AS [UnionAll11] UNION ALL SELECT 5 AS [C1] FROM (SELECT 1 AS X) AS [SingleRowTable15]) AS [UnionAll12]) AS [GroupBy1] WHERE [Extent1].[ContactID] IN ([GroupBy1].[A1], (CASE WHEN ([Limit1].[C1] IS NULL) THEN 0 ELSE [Limit2].[C1] END)) Based on this, is it possible that Linq2Entities provider indeed doesn't execute non-Linq and Linq-to-Object methods, but instead creates equivalent SQL statements for some of them ( and for others it throws an exception )? Thank you in advance

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  • getelementbyid does not work in firefox

    - by gaurab
    hi, this below mentioned code works perfect in internet explorer but not in firefox... i get an error in line in firefox: document.getElementById("supplier_no").value= values_array[0]; that getElementById returns null. how to solve the problem? var winName; //variable for the popup window var g_return_destination = null ; //variable to track where the data gets sent back to. // Set the value in the original pages text box. function f_set_home_value( as_Value ) { if (document.getElementById(g_return_destination[0]).name == "netbank_supplier_name_info" ) { //clear the old values for (selnum = 1; selnum <= 5; selnum++) { document.getElementById("expense_account"+selnum).value = ""; document.getElementById("expense_account_name"+selnum).value = ""; document.getElementById("expense_vat_flag"+selnum).value = "off"; document.getElementById("expense_vat_flag"+selnum).checked = ""; document.getElementById("expense_vat_amount"+selnum).value = ""; document.getElementById("expense_vat_code"+selnum).value = ""; document.getElementById("expense_period"+selnum).value = ""; document.getElementById("expense_date"+selnum).value = ""; if (selnum!=1) {//these are sometimes defaulted in, and in any case you will always have line1 document.getElementById("expense_more_dept"+selnum).value = ""; document.getElementById("expense_more_prj"+selnum).value = ""; document.getElementById("expense_more_subj"+selnum).value = ""; } document.getElementById("expense_amount"+selnum).value = ""; } var values_array = as_Value[0].split("!"); document.getElementById("supplier_no").value= values_array[0]; document.getElementById("supplier_bankAccount_no").value= values_array[1]; str = values_array[2] ; str = str.split(";sp;").join(" "); document.getElementById("default_expense_account").value= str; document.getElementById("expense_account1").value= str; document.getElementById("expense_more_sok1").disabled= false; str = values_array[3] ; str = str.split(";sp;").join(" "); document.getElementById("payment_term").value= str; strPeriod = calcPeriod(str,document.getElementById("due_date").value); document.getElementById("expense_period1").value = (strPeriod); strExpenseDate = calcExpenseDate(str,document.getElementById("due_date").value); document.getElementById("expense_date1").value = (strExpenseDate); str = values_array[4] ; str = str.split(";sp;").join(" "); document.getElementById("expense_account_name1").value= str; str = values_array[5] ; str = str.split(";sp;").join(" "); document.getElementById("expense_vat_code1").value= str; if (str == 0) { document.getElementById("expense_vat_flag1").checked= ''; document.getElementById("expense_vat_flag1").disabled= true; }else{ document.getElementById("expense_vat_flag1").checked= 'yes'; document.getElementById("expense_vat_flag1").value= 'on'; document.getElementById("expense_vat_flag1").disabled= false; } str = values_array[6] ; str = str.split(";sp;").join(" "); document.getElementById("supplier_name").value= str; var str = values_array[7]; str = str.split(";sp;").join(" "); str = str.split("&cr;").join("\r"); document.getElementById("netbank_supplier_name_info").value= str; strx = justNumberNF(document.getElementById("amount").value); document.all["expense_vat_amount1"].value = NetbankToDollarsAndCents(strx * (24/124)) ; document.getElementById("amount").value=NetbankToDollarsAndCents(strx); document.getElementById("expense_amount1").value = document.getElementById("amount").value; document.getElementById("expense_amount2").value = ''; document.getElementById("expense_account2").value= ''; //document.getElementById("expense_vat_flag2").value= ''; document.getElementById("expense_vat_amount2").value= ''; document.getElementById("expense_amount3").value = ''; document.getElementById("expense_account3").value= ''; //.getElementById("expense_vat_flag3").value= ''; document.getElementById("expense_vat_amount3").value= ''; document.getElementById("expense_amount4").value = ''; document.getElementById("expense_account4").value= ''; //document.getElementById("expense_vat_flag4").value= ''; document.getElementById("expense_vat_amount4").value= ''; document.getElementById("expense_amount5").value = ''; document.getElementById("expense_account5").value= ''; //document.getElementById("expense_vat_flag5").value= ''; document.getElementById("expense_vat_amount5").value= ''; str = values_array[8] ; str = str.split(";sp;").join(" "); if (str=="2"){ document.frmName.ButtonSelPeriodisering1.disabled=false; document.frmName.ButtonSelPeriodisering1.click(); } winName.close(); } } //Pass Data Back to original window function f_popup_return(as_Value) { var l_return = new Array(1); l_return[0] = as_Value; f_set_home_value(l_return); } function justNumberNF(val){ val = (val==null) ? 0 : val; // check if a number, otherwise try taking out non-number characters. if (isNaN(val)) { var newVal = parseFloat(val.replace(/[^\d\.\-]/g, '.')); // check if still not a number. Might be undefined, '', etc., so just replace with 0. return (isNaN(newVal) ? 0 : newVal); } // return 0 in place of infinite numbers. else if (!isFinite(val)) { return 0; } return val; }; function NetbankToDollarsAndCents(n) { var s = "" + Math.round(n * 100) / 100 ; var i = s.indexOf('.') ; if (i < 0) {return s + ",00" } ; var t = s.substring(0, i + 1) + s.substring(i + 1, i + 3) ; if (i + 2 == s.length) {t += "0"} ; return t.replace('.',',') ; }

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  • Handling inheritance with overriding efficiently

    - by Fyodor Soikin
    I have the following two data structures. First, a list of properties applied to object triples: Object1 Object2 Object3 Property Value O1 O2 O3 P1 "abc" O1 O2 O3 P2 "xyz" O1 O3 O4 P1 "123" O2 O4 O5 P1 "098" Second, an inheritance tree: O1 O2 O4 O3 O5 Or viewed as a relation: Object Parent O2 O1 O4 O2 O3 O1 O5 O3 O1 null The semantics of this being that O2 inherits properties from O1; O4 - from O2 and O1; O3 - from O1; and O5 - from O3 and O1, in that order of precedence. NOTE 1: I have an efficient way to select all children or all parents of a given object. This is currently implemented with left and right indexes, but hierarchyid could also work. This does not seem important right now. NOTE 2: I have tiggers in place that make sure that the "Object" column always contains all possible objects, even when they do not really have to be there (i.e. have no parent or children defined). This makes it possible to use inner joins rather than severely less effiecient outer joins. The objective is: Given a pair of (Property, Value), return all object triples that have that property with that value either defined explicitly or inherited from a parent. NOTE 1: An object triple (X,Y,Z) is considered a "parent" of triple (A,B,C) when it is true that either X = A or X is a parent of A, and the same is true for (Y,B) and (Z,C). NOTE 2: A property defined on a closer parent "overrides" the same property defined on a more distant parent. NOTE 3: When (A,B,C) has two parents - (X1,Y1,Z1) and (X2,Y2,Z2), then (X1,Y1,Z1) is considered a "closer" parent when: (a) X2 is a parent of X1, or (b) X2 = X1 and Y2 is a parent of Y1, or (c) X2 = X1 and Y2 = Y1 and Z2 is a parent of Z1 In other words, the "closeness" in ancestry for triples is defined based on the first components of the triples first, then on the second components, then on the third components. This rule establishes an unambigous partial order for triples in terms of ancestry. For example, given the pair of (P1, "abc"), the result set of triples will be: O1, O2, O3 -- Defined explicitly O1, O2, O5 -- Because O5 inherits from O3 O1, O4, O3 -- Because O4 inherits from O2 O1, O4, O5 -- Because O4 inherits from O2 and O5 inherits from O3 O2, O2, O3 -- Because O2 inherits from O1 O2, O2, O5 -- Because O2 inherits from O1 and O5 inherits from O3 O2, O4, O3 -- Because O2 inherits from O1 and O4 inherits from O2 O3, O2, O3 -- Because O3 inherits from O1 O3, O2, O5 -- Because O3 inherits from O1 and O5 inherits from O3 O3, O4, O3 -- Because O3 inherits from O1 and O4 inherits from O2 O3, O4, O5 -- Because O3 inherits from O1 and O4 inherits from O2 and O5 inherits from O3 O4, O2, O3 -- Because O4 inherits from O1 O4, O2, O5 -- Because O4 inherits from O1 and O5 inherits from O3 O4, O4, O3 -- Because O4 inherits from O1 and O4 inherits from O2 O5, O2, O3 -- Because O5 inherits from O1 O5, O2, O5 -- Because O5 inherits from O1 and O5 inherits from O3 O5, O4, O3 -- Because O5 inherits from O1 and O4 inherits from O2 O5, O4, O5 -- Because O5 inherits from O1 and O4 inherits from O2 and O5 inherits from O3 Note that the triple (O2, O4, O5) is absent from this list. This is because property P1 is defined explicitly for the triple (O2, O4, O5) and this prevents that triple from inheriting that property from (O1, O2, O3). Also note that the triple (O4, O4, O5) is also absent. This is because that triple inherits its value of P1="098" from (O2, O4, O5), because it is a closer parent than (O1, O2, O3). The straightforward way to do it is the following. First, for every triple that a property is defined on, select all possible child triples: select Children1.Id as O1, Children2.Id as O2, Children3.Id as O3, tp.Property, tp.Value from TriplesAndProperties tp -- Select corresponding objects of the triple inner join Objects as Objects1 on Objects1.Id = tp.O1 inner join Objects as Objects2 on Objects2.Id = tp.O2 inner join Objects as Objects3 on Objects3.Id = tp.O3 -- Then add all possible children of all those objects inner join Objects as Children1 on Objects1.Id [isparentof] Children1.Id inner join Objects as Children2 on Objects2.Id [isparentof] Children2.Id inner join Objects as Children3 on Objects3.Id [isparentof] Children3.Id But this is not the whole story: if some triple inherits the same property from several parents, this query will yield conflicting results. Therefore, second step is to select just one of those conflicting results: select * from ( select Children1.Id as O1, Children2.Id as O2, Children3.Id as O3, tp.Property, tp.Value, row_number() over( partition by Children1.Id, Children2.Id, Children3.Id, tp.Property order by Objects1.[depthInTheTree] descending, Objects2.[depthInTheTree] descending, Objects3.[depthInTheTree] descending ) as InheritancePriority from ... (see above) ) where InheritancePriority = 1 The window function row_number() over( ... ) does the following: for every unique combination of objects triple and property, it sorts all values by the ancestral distance from the triple to the parents that the value is inherited from, and then I only select the very first of the resulting list of values. A similar effect can be achieved with a GROUP BY and ORDER BY statements, but I just find the window function semantically cleaner (the execution plans they yield are identical). The point is, I need to select the closest of contributing ancestors, and for that I need to group and then sort within the group. And finally, now I can simply filter the result set by Property and Value. This scheme works. Very reliably and predictably. It has proven to be very powerful for the business task it implements. The only trouble is, it is awfuly slow. One might point out the join of seven tables might be slowing things down, but that is actually not the bottleneck. According to the actual execution plan I'm getting from the SQL Management Studio (as well as SQL Profiler), the bottleneck is the sorting. The problem is, in order to satisfy my window function, the server has to sort by Children1.Id, Children2.Id, Children3.Id, tp.Property, Parents1.[depthInTheTree] descending, Parents2.[depthInTheTree] descending, Parents3.[depthInTheTree] descending, and there can be no indexes it can use, because the values come from a cross join of several tables. EDIT: Per Michael Buen's suggestion (thank you, Michael), I have posted the whole puzzle to sqlfiddle here. One can see in the execution plan that the Sort operation accounts for 32% of the whole query, and that is going to grow with the number of total rows, because all the other operations use indexes. Usually in such cases I would use an indexed view, but not in this case, because indexed views cannot contain self-joins, of which there are six. The only way that I can think of so far is to create six copies of the Objects table and then use them for the joins, thus enabling an indexed view. Did the time come that I shall be reduced to that kind of hacks? The despair sets in.

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  • JoinDomainOrWorkgroup Method FJoinOptions help

    - by Ben
    Anyone have experience of using the JoinDomainOrWorkgroup Method of the Win32_ComputerSystem Class? I want to write a powershell script to join a machine to a domain. There may be an existing computer account for the machine, and if so I want to delete it and rejoin to the domain. I've already scripted the "search and destroy" part that will delete the computer account if it exists, but just noticed the FJoinOptions switches on Technet. Trouble is - they're a bit ambiguous. Does 4 (0x4) Deletes an account when a domain exists. mean it will delete the computer account if it already exists on the domain? Also, can you specify the computername you want to join the machine under with this method, or should you do a rename and then join the domain. Cheers, Ben NB - I've been using the guide at http://msdn.microsoft.com/en-us/library/aa392154(VS.85).aspx - not sure if there's a better resource out there.

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  • %HOMEPATH% Posh-Git error in Powershell, in ConEmu on Windows 7 64-bit

    - by atwright
    I am always getting the following error in Posh-Git in Powershell, in ConEmu on Windows 7 64-bit: Resolve-Path : Cannot find path 'C:\wamp\www\MobileApps\Backbone\%HOMEPATH%' because it does not exist. At D:\Users\Andy\Documents\WindowsPowerShell\Modules\posh-git\GitUtils.ps1:265 char:13 + $home = Resolve-Path (Invoke-NullCoalescing $Env:HOME ~) + ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ + CategoryInfo : ObjectNotFound: (C:\wamp\www\Mob...bone\%HOMEPATH%:String) [Resolve-Path], ItemNotFoundException + FullyQualifiedErrorId : PathNotFound,Microsoft.PowerShell.Commands.ResolvePathComma nd Join-Path : Cannot bind argument to parameter 'Path' because it is null. At D:\Users\Andy\Documents\WindowsPowerShell\Modules\posh-git\GitUtils.ps1:266 char:29 + Resolve-Path (Join-Path $home ".ssh\$File") -ErrorAction SilentlyContinue 2> ... + ~~~~~ + CategoryInfo : InvalidData: (:) [Join-Path], ParameterBindingValidationExc eption + FullyQualifiedErrorId : ParameterArgumentValidationErrorNullNotAllowed,Microsoft.Po werShell.Commands.JoinPathCommand Can anybody advise what might be wrong?

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  • Get mutually and non mutually existening Fields in same table in Two columns

    - by ranabra
    This is a question similar to another question I posted here but is a little different. I am trying to get a list of all instances of mutual and non-mutual existing Users. What I mean is that the returned result from the query will return a list of users along with their co-worker. It is similar to the question here, but the difference is that non mutual users will be returned too and with out the "duplicity" mutually existing users return in the list (See image below in-order simplify it all). I took the original answer from Thomas (Thanx again Thomas) Select D1.u_username, U1.Permission, U1.Grade, D1.f_username, U2.Permission, U2.Gradefrom tblDynamicUserList As D1    Join tblDynamicUserList As D2        On D2.u_username = D1.f_username            And D2.f_username = D1.u_username    Join tblUsers As U1        On U1.u_username = D1.u_username    Join tblUsers As U2        On U2.u_username = D2.u_username and after some several trials I commented out 2 lines (Below). The returned result are exactly as described in the beginning of this question, but with the "duplicity" returned by mutually existing users in the table. How can I eliminate this duplicity? Select D1.u_username, U1.Permission, U1.Grade, D1.f_username, U2.Permission, U2.Gradefrom tblDynamicUserList As D1    Join tblDynamicUserList As D2        On D2.u_username = D1.f_username            /* And D2.f_username = D1.u_username */    Join tblUsers As U1        On U1.u_username = D1.u_username    Join tblUsers As U2        On U2.u_username = D2.u_username /* WHERE D1.U_userName < D1.f_username */ *Screenshot that hopefully helps explain it all. Database is SQL 2005. Many thanx in advance

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  • Network Path not found while joining Active Directory

    - by Chiggins
    So I have an Amazon EC2 box running Windows Server 2008 with Active Directory installed on it. I also have a Windows 7 virtual machine, which is set to use the Active Directory box as its DNS and WINS server. I'm trying to join the virtual machine to the domain, and I'm asked for authentication. I give authentication, wait a minute, and I get an error saying: The following error occurred attempting to join the domain "ad.chigs.me": The network path was not found. How can I fix this so that I'll be able to join the domain?

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  • Samba Server as domain controller

    - by garden air
    I am using centos 6.2 on my system. I want to make samba server as PDC to join the clients computers operating systems i.e xp,windows 7 and share their files & directories.Currently I have 200 PC running both windows xp and win 7. The question I want to as is does samba 3.5.10 has a capacity to join 200 computers as a domain controller & authenticate the users ? thanks garden Thanks for your guidence.Well at the moment I am using CentOS 6.2 and samba is installed using yum command. It is amazing that I can join samba as a domain comtroller of 200 clients machines. By the way what is samba maximum limit for joining number of clients PC's.

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  • MySQL query works in PHPMyAdmin but not PHP

    - by Su4p
    I do not understand what's happening. I have a query in PHP who crashes -with a strange error-. When I copy/paste the exact same request in PHPMyAdmin it works as expected. What am I doing wrong here ? SELECT oms_patient.id, oms_patient.date, oms_patient.date_modif, date_modif, AES_DECRYPT(nom,"xxxxx") AS "Nom", AES_DECRYPT(prenom,"xxxxx") AS "Prénom usuel", DATE_FORMAT(ddn, "%d/%m/%Y") AS "Date de naissance", villeNaissance AS "Lieu de naissance (ville)", CONCAT(oms_departement.libelle,"(",id_departement,")") AS "Lieu de vie", CONCAT(oms_pays.libelle,"(",id_pays,")") AS "Pays", CONCAT(patientsexe.libelle,"(",id_sexe,")") AS "Sexe", CONCAT(patientprofession.libelle,"(",id_profession,")") AS "Profession", IF(asthme>0,"Oui","Non") AS "Asthme", IF(rhinite>0,"Oui","Non") AS "Rhinite", IF(bcpo>0,"Oui","Non") AS "BPCO", IF(insuffisanceResp>0,"Oui","Non") AS "Insuffisance respiratoire chronique", IF(chirurgieOrl>0,"Oui","Non") AS "Chirurgie ORL du ronflement", IF(autreChirurgie>0,"Oui","Non") AS "Autre chirurgie ORL", IF(allergies>0,"Oui","Non") AS "Allergies", IF(OLD>0,"Oui","Non") AS "OLD", IF(hypertensionArterielle>0,"Oui","Non") AS "Hypertension artérielle", IF(infarctusMyocarde>0,"Oui","Non") AS "Infarctus du myocarde", IF(insuffisanceCoronaire>0,"Oui","Non") AS "Insuffisance coronaire", IF(troubleRythme>0,"Oui","Non") AS "Trouble du rythme", IF(accidentVasculaireCerebral>0,"Oui","Non") AS "Accident vasculaire cérébral", IF(insuffisanceCardiaque>0,"Oui","Non") AS "Insuffisance cardiaque", IF(arteriopathie>0,"Oui","Non") AS "Artériopathie", IF(tabagismeActuel>0,"Oui","Non") AS "Tabagisme actuel", CONCAT(nbPaquetsActuel," ","PA") AS "", IF(tabagismeAncien>0,"Oui","Non") AS "Tabagisme ancien", CONCAT(nbPaquetsAncien," ","PA") AS "", IF(alcool>0,"Oui","Non") AS "Alcool (conso régulière)", IF(refluxGastro>0,"Oui","Non") AS "Reflux gastro-oesophagien", IF(glaucome>0,"Oui","Non") AS "Glaucome", IF(diabete>0,"Oui","Non") AS "Diabète", CONCAT(patienttypeDiabete.libelle,"(",id_typeDiabete,")") AS "", IF(hypercholesterolemie>0,"Oui","Non") AS "Hypercholestérolémie", IF(hypertriglyceridemie>0,"Oui","Non") AS "Hypertriglycéridémie", IF(dysthyroidie>0,"Oui","Non") AS "Dysthyroïdie", IF(depression>0,"Oui","Non") AS "Dépression", IF(sedentarite>0,"Oui","Non") AS "Sédentarité", IF(syndromeDApneesSommeil>0,"Oui","Non") AS "SAS", IF(obesite>0,"Oui","Non") AS "Obésité", IF(dysmorphieFaciale>0,"Oui","Non") AS "Dysmorphie faciale", TextObservations AS "", id_user FROM oms_patient LEFT JOIN oms_departement ON oms_departement.id = id_departement LEFT JOIN oms_pays ON oms_pays.id = id_pays LEFT JOIN patientsexe ON patientsexe.id = id_sexe LEFT JOIN patientprofession ON patientprofession.id = id_profession LEFT JOIN patienttypeDiabete ON patienttypeDiabete.id = id_typeDiabete WHERE oms_patient.id=1 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'small"(conso régulière)", IF(refluxGastro0,"Oui","Non") as "Reflux ga' at line 1 "near 'small" <-- where is small o_O The PHP code isn't really relevant cause you won't see a lot. $db = mysql_connect(); mysql_select_db();//TODO SWITCH TO PDO mysql_query("SET NAMES UTF8"); $fields = $form->getFields($form); $settingsForm = $form->getSettings(); $sql = 'SELECT oms_patient.id,oms_patient.date,oms_patient.date_modif,'; foreach ($fields as $field) { if (!$field->isMultiSelect()) { $field->select_full(&$sql, 'oms_patient', null); } } if (isset($settingsForm['linkTo'])) { $idLinkTo = 'id_' . str_replace('oms_', '', $settingsForm['linkTo']); $sql .= $idLinkTo; } $sql.=' FROM oms_patient'; foreach ($fields as $field) { if (!$field->isMultiSelect() && $field->getTable('oms_patient')) { $sql .=' LEFT JOIN ' . $field->getTable('oms_patient') . ' ON ' . $field->getTable('oms_patient') . '.id = '.$field->getFieldName().' '; } } $sql.=' where oms_patient.id=' . $this->m_settings['e']; $result = mysql_query($sql) or die('Erreur SQL !<br>' . $sql . '<br>' . mysql_error()); $data = mysql_fetch_assoc($result); var_dump of $sql string(2663) "SELECT oms_patient.id,oms_patient.date,oms_patient.date_modif,date_modif,AES_DECRYPT(nom,"xxxxx") as "Nom",AES_DECRYPT("prenom","xxxxx") as "Prénom usuel",DATE_FORMAT(ddn, "%d/%m/%Y") as "Date de naissance",villeNaissance as "Lieu de naissance (ville)",CONCAT(oms_departement.libelle,"(",id_departement,")") as "Lieu de vie",CONCAT(oms_pays.libelle,"(",id_pays,")") as "Pays",CONCAT(patientsexe.libelle,"(",id_sexe,")") as "Sexe",CONCAT(patientprofession.libelle,"(",id_profession,")") as "Profession", IF"... can't go further to see what is in the output after the "..." <-- if you have an idea

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  • Need help to properly remove duplicates in NHibernate

    - by Michael D. Kirkpatrick
    Here is the problem I am having. I have a database with over 100 records in it. I am paging through the data to get 9 results at a time. When I added a check to see if items are active, it caused the results to start doubling up. A little background: "Product" is the actual product line "ProductSkus" are the actual products that exist in the product line When there is more then 1 ProductSku within Product, it causes a duplicate entry to be returned. See the NHibernate Query below: result = this.Session.CreateCriteria<Model.Product>() .Add(Expression.Eq("IsActive", true)) .AddOrder(new Order("Name", true)) .SetFirstResult(indexNumber).SetMaxResults(maxNumber) // This part of the query duplicates the products .CreateAlias("ProductSkus", "ProdSkus", JoinType.InnerJoin) .Add(Expression.Eq("ProdSkus.IsActive", true)) .CreateAlias("ProductToSubcategory", "ProdToSubcat") .CreateAlias("ProdToSubcat.ProductSubcategory", "ProdSubcat") .Add(Expression.Eq("ProdSubcat.ID", subCatId)) // This part takes out the duplicate products - Removes too many items... // Turns out that with .SetFirstResult(indexNumber).SetMaxResults(maxNumber) // it gets 9 records back then the duplicates are removed. // Example: // Total Records over 100 // Max = 9 // 4 Duplicates removed // Yields 5 records when there should be 9 // Why??? This line is ran in NHibernate on the data after it has been extracted from the SQL server. .SetResultTransformer(new NHibernate.Transform.DistinctRootEntityResultTransformer()) .List<Model.Product>(); I added the DistinctRootEntityResultTransformer to clean up the duplicates. The problem is that it pulls 9 records back that contains duplicates. DistinctRootEntityResultTransformer then cleans up the duplicates in the 9 records. I am basically needing a distinct statement to be ran on the SQL server to begin with. However, distinct on SQL is not going to work since NHibernate by default wants to add every field from every table in the select part of the statement. I am only using the fields that belong to the root table to begin with (Model.Product). If I can tell NHibernate to not add the fields to the joined tables into the select part of the statement along with adding Distinct, it would work. I use NHibernare Profiler to see the actual query: SELECT top 9 this_.ID as ID351_3_, this_.Name as Name351_3_, this_.Description as Descript3_351_3_, this_.IsActive as IsActive351_3_, this_.ManufacturerID as Manufact5_351_3_, prodskus1_.ID as ID373_0_, prodskus1_.Description as Descript2_373_0_, prodskus1_.PartNumber as PartNumber373_0_, prodskus1_.Price as Price373_0_, prodskus1_.IsKit as IsKit373_0_, prodskus1_.IsActive as IsActive373_0_, prodskus1_.IsFeaturedProduct as IsFeatur7_373_0_, prodskus1_.DateAdded as DateAdded373_0_, prodskus1_.Weight as Weight373_0_, prodskus1_.TimesViewed as TimesVi10_373_0_, prodskus1_.TimesOrdered as TimesOr11_373_0_, prodskus1_.ProductID as ProductID373_0_, prodskus1_.OverSizedBoxID as OverSiz13_373_0_, prodtosubc2_.ID as ID362_1_, prodtosubc2_.MasterSubcategory as MasterSu2_362_1_, prodtosubc2_.ProductID as ProductID362_1_, prodtosubc2_.ProductSubcategoryID as ProductS4_362_1_, prodsubcat3_.ID as ID352_2_, prodsubcat3_.Name as Name352_2_, prodsubcat3_.ProductCategoryID as ProductC3_352_2_, prodsubcat3_.ImageID as ImageID352_2_, prodsubcat3_.TriggerShow as TriggerS5_352_2_ FROM Product this_ inner join ProductSku prodskus1_ on this_.ID = prodskus1_.ProductID and (prodskus1_.IsActive = 1) inner join ProductToSubcategory prodtosubc2_ on this_.ID = prodtosubc2_.ProductID inner join ProductSubcategory prodsubcat3_ on prodtosubc2_.ProductSubcategoryID = prodsubcat3_.ID WHERE this_.IsActive = 1 /* @p0 */ and prodskus1_.IsActive = 1 /* @p1 */ and prodsubcat3_.ID = 3 /* @p2 */ ORDER BY this_.Name asc If I hand modify the query and run it directly on the SQL server I get the result set I want (I removed all the extra fields in the select section and added DISTINCT): SELECT DISTINCT top 9 this_.ID as ID351_3_, this_.Name as Name351_3_, this_.Description as Descript3_351_3_, this_.IsActive as IsActive351_3_, this_.ManufacturerID as Manufact5_351_3_, FROM Product this_ inner join ProductSku prodskus1_ on this_.ID = prodskus1_.ProductID and (prodskus1_.IsActive = 1) inner join ProductToSubcategory prodtosubc2_ on this_.ID = prodtosubc2_.ProductID inner join ProductSubcategory prodsubcat3_ on prodtosubc2_.ProductSubcategoryID = prodsubcat3_.ID WHERE this_.IsActive = 1 /* @p0 */ and prodskus1_.IsActive = 1 /* @p1 */ and prodsubcat3_.ID = 3 /* @p2 */ ORDER BY this_.Name asc The big question I now must ask is... What must I change in the NHibernate Query to ultimately get the exact same result? Thanks in advance.

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  • Network Path not found while joining Active Directory

    - by Chiggins
    So I have an Amazon EC2 box running Windows Server 2008 with Active Directory installed on it. I also have a Windows 7 virtual machine, which is set to use the Active Directory box as its DNS and WINS server. I'm trying to join the virtual machine to the domain, and I'm asked for authentication. I give authentication, wait a minute, and I get an error saying: The following error occurred attempting to join the domain "ad.chigs.me": The network path was not found. How can I fix this so that I'll be able to join the domain? Thanks!

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