FIX: You may receive incorrect results when you run a Transact-SQL query in SQL Server 2000

• The issues that are fixed by this hotfix package
• The prerequisites for applying this hotfix package
• Whether you must restart the computer after you apply this hotfix package
• Whether this hotfix package is replaced by any other hotfix package
• Whether you must make any registry changes after you apply this hotfix package
• The files that are contained in this hotfix package

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• The query contains an inner join that has a one-to-one key relationship.
• The query contains an aggregate expression or a DISTINCT keyword.
• The GROUP BY clause appears in the Showplan.
• The query contains an outer join.
• The outer join is simplified out of the query because no columns from the outer join table are referenced elsewhere in the query.

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• The system uses the Advanced Micro Devices (AMD) AMD64 processor architecture or the Intel Extended Memory 64 Technology (EM64T) processor architecture.
  
  Note This issue does not occur on systems that use the Intel Itanium processor architecture.
• The system is running a 64-bit version of the Microsoft Windows Server operating system.
• The system is running a 32-bit version of SQL Server 2000.

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Hotfix information

Prerequisites

Restart information

Hotfix file information

SQL Server 2000, 32-bit versions

SQL Server 2000, Itanium architecture version

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Steps to reproduce the problem

Create the test environment

USE Northwind
GO

CREATE TABLE [dbo].[Participant] (
	[ProgramUserId] [int] NOT NULL ,
	[ProjectId] [int] NOT NULL 
) ON [PRIMARY]
GO
CREATE TABLE [dbo].[Project] (
	[ProjectId] [int] NOT NULL ,
	[ProjectHdrId] [int] NOT NULL ,
	[CustomerId] [int] NOT NULL 
) ON [PRIMARY]
GO
CREATE TABLE [dbo].[ProjectHdr] (
	[ProjectHdrId] [int] NOT NULL
) ON [PRIMARY]
GO
ALTER TABLE [dbo].[Participant] WITH NOCHECK ADD 
	CONSTRAINT [pkParticipant_ParticipantId] PRIMARY KEY  CLUSTERED 
	(
		[ProgramUserId],
		[ProjectId]
	)  ON [PRIMARY] 
GO
ALTER TABLE [dbo].[Project] WITH NOCHECK ADD 
	CONSTRAINT [pkProject_ProjectId] PRIMARY KEY  CLUSTERED 
	(
		[ProjectId]
	)  ON [PRIMARY] 
GO
ALTER TABLE [dbo].[ProjectHdr] WITH NOCHECK ADD 
	CONSTRAINT [pkProjectHdr_ProjectHdrId] PRIMARY KEY  CLUSTERED 
	(
		[ProjectHdrId]
	)  ON [PRIMARY] 
GO
 CREATE  UNIQUE  INDEX [ixProject_CustomerId_Unique] ON [dbo].[Project]([CustomerId], [ProjectHdrId]) ON [PRIMARY]
GO
insert into Participant(ProgramUserId,ProjectId) values('3','8');
insert into Participant(ProgramUserId,ProjectId) values('3','74');
insert into Participant(ProgramUserId,ProjectId) values('3','1004');
insert into Participant(ProgramUserId,ProjectId) values('5','1003');
insert into Participant(ProgramUserId,ProjectId) values('7','8');
insert into Participant(ProgramUserId,ProjectId) values('7','74');
insert into Participant(ProgramUserId,ProjectId) values('7','1004');
insert into Participant(ProgramUserId,ProjectId) values('8','1003');
insert into Participant(ProgramUserId,ProjectId) values('8','1004');
GO

insert into Project (ProjectId,ProjectHdrId,CustomerId) values('1','1','1');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('2','2','2');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('7','15','5');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('8','26','5');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('71','27','52');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('72','3','52');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('73','3','51');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('74','15','51');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('1003','1002','5');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('1004','1001','5');
insert into Project (ProjectId,ProjectHdrId,CustomerId) values('1014','1001','51');
GO

insert into ProjectHdr(ProjectHdrId) values('1');
insert into ProjectHdr(ProjectHdrId) values('2');
insert into ProjectHdr(ProjectHdrId) values('3');
insert into ProjectHdr(ProjectHdrId) values('15');
insert into ProjectHdr(ProjectHdrId) values('26');
insert into ProjectHdr(ProjectHdrId) values('27');
insert into ProjectHdr(ProjectHdrId) values('1001');
insert into ProjectHdr(ProjectHdrId) values('1002');
GO

Analyze the problem

1. Run the following SQL query.

   SELECT PH.ProjectHdrId, P.ProjectId
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5

   As expected, the query returns eight rows. Some rows are duplicates. Therefore, there are four distinct rows.
2. Add the DISTINCT keyword, and then run the query again.

   dbcc freeproccache
   
   SELECT DISTINCT PH.ProjectHdrId, P.ProjectId
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5
   

   The result of this query should be four rows. However, only one row is returned.

Work around the problem

1. The following workaround returns the correct result set for the JOIN query. This query uses a temporary table and selected distinct values from the query.

   dbcc freeproccache
   
   SELECT PH.ProjectHdrId, P.ProjectId
   INTO #temp
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5
   SELECT * from #temp
   SELECT DISTINCT * from #temp
   DROP TABLE #temp
   

   If the SELECT statement in this query does not have the DISTINCT keyword, the query returns eight rows. If the SELECT statement in this query has the DISTINCT keyword, this query returns four rows.
2. The following query adds a "dummy condition" on the Participant table. Run this query.

   dbcc freeproccache
   
   SELECT DISTINCT PH.ProjectHdrId, P.ProjectId
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5 AND (PP.ProjectId IS NOT NULL OR PP.ProjectId IS NULL)
   

   Notice that the query returns four rows.
3. The following query changes the order of the equi-join and enforces this order. Run this query.

   dbcc freeproccache
   
   SELECT DISTINCT PH.ProjectHdrId, P.ProjectId
   FROM Project AS P JOIN ProjectHdr AS PH ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5 OPTION (FORCE ORDER)
   

   Notice that the query returns four rows.
4. The following query drops the index Project.ixProject_CustomerId_Unique. Run this query.

   drop index project.ixProject_CustomerId_Unique
   
   dbcc freeproccache
   
   SELECT DISTINCT PH.ProjectHdrId, P.ProjectId
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5

   Notice that the query returns four rows.

Reproduce the problem

1. If you re-create the index Project.ixProject_CustomerId_Unique, the problem occurs again. Run the following code to re-create the index.

   create unique index ixProject_CustomerId_Unique on project (CustomerId, ProjectHdrId)

2. Run the following query.

   dbcc freeproccache
   
   SELECT DISTINCT PH.ProjectHdrId,  P.ProjectId
   FROM ProjectHdr AS PH JOIN Project AS P ON P.ProjectHdrId = PH.ProjectHdrId
   LEFT OUTER JOIN Participant AS PP ON PP.ProjectId = P.ProjectId
   WHERE P.CustomerId = 5

   Notice that only one record is returned. However, if you run the query without the DISTINCT keyword in the SELECT statement, the query returns eight rows.

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