SQL中的時間相加減

select datediff(minute,'2012-05-31 08:03','2012-06-01 08:05') --算出兩個時間相差幾分鐘

SELECT DATEADD(minute,5,'2012-05-01 08:03:01')

select dateAdd(minute,19,'2008-10-25 08:05:31')

select dateAdd(hour,19,'2008-10-25 08:05:31')

select dateAdd(second,19,'2008-10-25 08:05:31')

select dateAdd(day,19,'2008-10-25 08:05:31')

select dateAdd(month,19,'2008-10-25 08:05:31')

select dateAdd(year,19,'2008-10-25 08:05:31')

select cast('2008-10-25 08:05:31'as datetime) +cast(5 as float)/1440

--float 為1時 ，相當與一天