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SQL Server add job agent T-SQL script issue

I am using SQL Server 2008 Enterprise. I have created a very simple test SQL Server Agent job which just print out a string. And then I add a schedule to let the job run once a day.

Then I select from SQL Server Management Studio => Jobs => Select the test job => Script Job as => Drop and Create To, here is the generated scripts. My confusion is, when (at the bottom of the script) calls msdb.dbo.sp_add_jobschedule, there is an input parameter provided as an input parameter for this store procedure called @schedule_uid, but from MSDN, this parameter is an output parameter, and not an input parameter. Any ideas why is wrong and why such mismatching?

Here is MSDN page and the script generated from SQL Server Management Studio,

http://msdn.microsoft.com/en-us/library/ms366342.aspx

USE [msdb]
GO

/****** Object:  Job [TestJob]    Script Date: 08/05/2009 14:21:59 ******/
IF  EXISTS (SELECT job_id FROM msdb.dbo.sysjobs_view WHERE name = N'TestJob')
EXEC msdb.dbo.sp_delete_job @job_id=N'dd219677-7584-43de-b14c-b583b6ff619e', @delete_unused_schedule=1
GO

USE [msdb]
GO

/****** Object:  Job [TestJob]    Script Date: 08/05/2009 14:21:59 ******/
BEGIN TRANSACTION
DECLARE @ReturnCode INT
SELECT @ReturnCode = 0
/****** Object:  JobCategory [[Uncategorized (Local)]]]    Script Date: 08/05/2009 14:21:59 ******/
IF NOT EXISTS (SELECT name FROM msdb.dbo.syscategories WHERE name=N'[Uncategorized (Local)]' AND category_class=1)
BEGIN
EXEC @ReturnCode = msdb.dbo.sp_add_category @class=N'JOB', @type=N'LOCAL', @name=N'[Uncategorized (Local)]'
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback

END

DECLARE @jobId BINARY(16)
EXEC @ReturnCode =  msdb.dbo.sp_add_job @job_name=N'TestJob', 
        @enabled=1, 
        @notify_level_eventlog=3, 
        @notify_level_email=0, 
        @notify_level_netsend=0, 
        @notify_level_page=0, 
        @delete_level=0, 
        @description=N'No description available.', 
        @category_name=N'[Uncategorized (Local)]', 
        @owner_login_name=N'Administrator', @job_id = @jobId OUTPUT
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback
/****** Object:  Step [TestStep1]    Script Date: 08/05/2009 14:21:59 ******/
EXEC @ReturnCode = msdb.dbo.sp_add_jobstep @job_id=@jobId, @step_name=N'TestStep1', 
        @step_id=1, 
        @cmdexec_success_code=0, 
        @on_success_action=1, 
        @on_success_step_id=0, 
        @on_fail_action=2, 
        @on_fail_step_id=0, 
        @retry_attempts=0, 
        @retry_interval=0, 
        @os_run_priority=0, @subsystem=N'TSQL', 
        @command=N'PRINT "Here is a test job"', 
        @database_name=N'master', 
        @flags=0
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback
EXEC @ReturnCode = msdb.dbo.sp_update_job @job_id = @jobId, @start_step_id = 1
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback
EXEC @ReturnCode = msdb.dbo.sp_add_jobschedule @job_id=@jobId, @name=N'TestJobSchedule', 
        @enabled=1, 
        @freq_type=8, 
        @freq_interval=1, 
        @freq_subday_type=1, 
        @freq_subday_interval=0, 
        @freq_relative_interval=0, 
        @freq_recurrence_factor=1, 
        @active_start_date=20090805, 
        @active_end_date=99991231, 
        @active_start_time=0, 
        @active_end_time=235959, 
        @schedule_uid=N'68d6209d-86a7-48a4-86ea-51322f279a2d'
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback
EXEC @ReturnCode = msdb.dbo.sp_add_jobserver @job_id = @jobId, @server_name = N'(local)'
IF (@@ERROR <> 0 OR @ReturnCode <> 0) GOTO QuitWithRollback
COMMIT TRANSACTION
GOTO EndSave
QuitWithRollback:
    IF (@@TRANCOUNT > 0) ROLLBACK TRANSACTION
EndSave:

GO
like image 727
George2 Avatar asked Mar 20 '26 02:03

George2


1 Answers

My guess would be that the @schedule_uid can either be defined and returned by SQL Server upon job creation (the OUTPUT case), or it can be set manually (obviously undocumented).

My guess would further be that the undocumented INPUT case is a means to maintaining the original uid during drop and recreate.

like image 84
Tomalak Avatar answered Mar 21 '26 14:03

Tomalak



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