时间的计算问题？

Question

--是这个意思吧?select id,使用的分钟数
=isnull(datediff(minute,downtime1,uptime1),0)
+isnull(datediff(minute,downtime2,uptime2),0)
+isnull(datediff(minute,downtime3,uptime3),0)
from 表

Answer 1

datediff(minute,'2004-1-1 '+isnull(uptime1,'00:00:00'),'2004-1-1 '+isnull(downtime1,'00:00:00'))
+datediff(minute,'2004-1-1 '+isnull(uptime2,'00:00:00'),'2004-1-1 '+isnull(downtime2,'00:00:00'))
+datediff(minute,'2004-1-1 '+isnull(uptime2,'00:00:00'),'2004-1-1 '+isnull(downtime2,'00:00:00'))

Answer 2

select id,
       datediff(minute,downtime1 ,  uptime1 )+
       datediff(minute,downtime2 ,  uptime2 )+
       datediff(minute,downtime3 ,  uptime3 ) as dtime
   from tablename

Answer 3

null+null=null，
所以zjcxc(邹建)的对！

Answer 4

多谢！这样就行了，不知怎么将那个＜NULL＞的数值加上去而已，
依然那个方法不行的，如果UPtimer 的数值为＜NULL＞的话那么结果加上来也还是为＜null＞的