DATEDIFF(hour,'2009-07-28 14:33:08.887',GETDATE())
DATEDIFF(day,'2009-07-28 14:33:08.887',GETDATE())
--错误
declare @timetype varchar(10)
set @timetype='hour'
select DATEDIFF(@timetype,'2009-07-28 14:33:08.887',GETDATE())我想 将 datediff 中 的hour,或day 作为一个变量传进来。请大家 赐教!!
DATEDIFF(day,'2009-07-28 14:33:08.887',GETDATE())
--错误
declare @timetype varchar(10)
set @timetype='hour'
select DATEDIFF(@timetype,'2009-07-28 14:33:08.887',GETDATE())我想 将 datediff 中 的hour,或day 作为一个变量传进来。请大家 赐教!!
declare @timetype varchar(10)
declare @SQl varchar(1000)
set @timetype='hour'
Set @sql = 'select DATEDIFF('+@timetype+',''2009-07-28 14:33:08.887'',GETDATE())'
exce(@sql)
declare @timetype varchar(10)
declare @SQl varchar(1000)
set @timetype='hour'
Set @sql = 'select DATEDIFF('+@timetype+',''2009-07-28 14:33:08.887'',GETDATE())'
exec(@sql)
SELECT @timetype=DATEDIFF(HOUR,'2009-07-28 14:33:08.887',GETDATE())
set @timetype='hour'
set @s='select @Dt=DATEDIFF('+@timetype+',''2009-07-28 14:33:08.887'',GETDATE())'
exec sp_executesql @s,N'@Dt bigint output',@Dt output
select @Dt