datetime类型如何计算相差的时间？

[Fabulous]

例如:
2010-08-30与1988-4-6相差多少天？多少月？
怎么写代码啊？

[bancxc]

datediff(day,startdate,enddate)
datediff(month,startdate,enddate)

[obuntu]

select DATEDIFF(month ,'1988-4-6','2010-08-30'),
	    DATEDIFF(day  ,'1988-4-6','2010-08-30')

[百年树人]

if object_id('[tb]') is not null drop table [tb]
go
create table [tb]([col1] datetime,[col2] datetime)
insert [tb]
select '2010-08-30','1988-4-6'
go

select
  datediff(dd,col2,col1) as 相差天数,
  datediff(mm,col2,col1) as 相差月数
from
  tb 

/**
相差天数        相差月数
----------- -----------
8181        268

(1 行受影响)
**/

[jaydom]

select datediff(day,'1988-4-6','2010-08-30')--8181 days

select datediff(month,'1988-4-6','2010-08-30')--268 months

[Fabulous]

谢谢大家， 鲁迅先生的稍稍复杂了一些，我刚学这个看起来吃力。