SQL 时间差计算

显示全部楼层 · 2009-7-11 10:20:28

同一个ID有可能存在多次登陆和登出用嵌套--> --> (Roy)生成测试数据 declare @T table([UserID] int,[OnP] nvarchar(3),[OnTime] Datetime)Insert @Tselect 1,N'on','2007-7-1 7:00' union allselect 1,N'off','2007-7-1 8:00' union allselect 2,N'on','2007-8-9 12:45' union allselect 3,N'on','2007-8-9 13:50' union allselect 2,N'off','2007-8-9 16:1...

千问 · 2009-7-11 10:20:28

把OnP='off'的放到一个临时表中select * into #AA from A where OnP='off'把没有离开时间的用户加到临时表中insert into #AA (select * from A where UserID not in (select UserID from #AA))把临时表中的上面加的改一下登陆状态和时...

千问 · 2009-7-11 10:20:28

用datediff函数就搞定了...

千问 · 2009-7-11 10:20:28

有意思我试试密切关注此问题...

千问 · 2009-7-11 10:20:28

1.select a.userID,dateDiff(Hour,a.onTime,isnull(b.Ontime,getdate())) as onlineTime from A aleft joinA b on a.UserID=b.UserID and a.onp='On' and b.onp='OFF'2.select a.userID,s...