SQL Server 延时执行

参考资料:SQL Server 2008中SQL之WaitFor在SQL Server 2005以上版本中,在一个增强的WaitFor命令,其作用可以和一个job相当。但使用更加简捷。

语法为:

WAITFOR 
{
DELAY 'time_to_pass'
| TIME 'time_to_execute'
| [ ( receive_statement ) | ( get_conversation_group_statement ) ]
[ , TIMEOUT timeout ]
}

--示例1: 等待n秒后执行

Declare @t1 Varchar(8)
Set @t1 = Convert(Varchar,DateAdd(ss, @Secnod, 0),108)
WAITFOR DELAY @t1   --等待n秒


--示例2:等待n秒后执行
Declare @t1 DateTime, @n int
Select @t1 = GetDate(), @n = 0
While DateDiff(s,@t1,GetDate())<@Second 
Set @n = @n + 1

--示例3:以下示例在晚上 10:20 (22:20) 执行存储过程 sp_update_job。
USE msdb;
EXECUTE sp_add_job @job_name = 'TestJob';
BEGIN
WAITFOR TIME '22:20';
EXECUTE sp_update_job @job_name = 'TestJob',
@new_name = 'UpdatedJob';
END;


--示例4:以下示例在两小时的延迟后执行存储过程。注意:Delay最多不超过24小时
BEGIN
WAITFOR DELAY '02:00';
EXECUTE sp_helpdb;

END;

以下示例显示如何对 WAITFOR DELAY 选项使用局部变量。将创建一个存储过程,该过程将等待可变的时间段,然后将经过的小时、分钟和秒数信息返回给用户。

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USE AdventureWorks2008R2;
GO
IF OBJECT_ID ( ' dbo.TimeDelay_hh_mm_ss ' , ' P ' ) IS NOT NULL
DROP PROCEDURE dbo.TimeDelay_hh_mm_ss;
GO
CREATE PROCEDURE dbo.TimeDelay_hh_mm_ss
(
@DelayLength char ( 8 ) = ' 00:00:00 '
)
AS
DECLARE @ReturnInfo varchar ( 255 )
IF ISDATE ( ' 2000-01-01 ' + @DelayLength + ' .000 ' ) = 0
BEGIN
SELECT @ReturnInfo = ' Invalid time ' + @DelayLength
+ ' ,hh:mm:ss, submitted. ' ;
-- This PRINT statement is for testing, not use in production.
PRINT @ReturnInfo
RETURN ( 1 )
END
BEGIN
WAITFOR DELAY @DelayLength
SELECT @ReturnInfo = ' A total time of ' + @DelayLength + ' ,
hh:mm:ss, has elapsed! Your time is up.
'
-- This PRINT statement is for testing, not use in production.
PRINT @ReturnInfo ;
END ;
GO
/* This statement executes the dbo.TimeDelay_hh_mm_ss procedure. */
EXEC TimeDelay_hh_mm_ss ' 00:00:10 ' ;
GO
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执行结果:
A total time of 00:00:10, in hh:mm:ss, has elapsed.Your time is up.

小结:这是一种轻巧的解决方案。当你没有权限指定job时,可以考虑用WaitFor语句。

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