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我認爲將它正確完成是微不足道的。帶有存儲過程的SQL SERVER 2008最終選擇命令
在屏幕截圖下面的每個用戶(用戶ID)有兩個結果(這是一種重複的)
如何解決查詢來獲取每個用戶
(每個用戶只有一個結果可以有2組「TimeIn」和「TimeOut」活動)
所以如果給定的用戶確實有第二個「入口」,但沒有離開 我只需要第一個封閉的入口/離開+第二個入口/仍然工作
這裏是存儲過程
create table #tmp (tId int, UserId int,
TimeIn1 smalldatetime, [TimeOut1] smalldatetime,
TimeIn2 smalldatetime, [TimeOut2] smalldatetime, tId2 int,
ActiveDate smalldatetime, ReasonID int, Name nvarchar(100), ReasonType nvarchar(100),
TotalMins int)
insert into #tmp (tId, UserId, TimeIn1, TimeOut1, ActiveDate, ReasonID, Name, ReasonType)
SELECT
t1.tId, t1.UserId, t1.TimeIn, t1.[TimeOut], t1.ActiveDate, t1.ReasonID, tblCustomers.name,tblTimeReas.ReasonType
FROM tblTime t1
inner join tblTimeReas on t1.ReasonID = tblTimeReas.ReasonID
inner join tblCustomers on t1.UserId=tblCustomers.custID
where (t1.userid in (select custID from tblCustomers where Classification =35))
and (DATEPART(DAY,t1.timein)= DATEPART(DAY,GETDATE()))
and (DATEPART(MONTH,t1.timein)= DATEPART(MONTH,GETDATE()))
and (DATEPART(YEAR,t1.timein)= DATEPART(YEAR,GETDATE()))
update #tmp
set tId2 = (select top 1 tId from
tblTime t2 where (userid in (select custID from tblCustomers where Classification =35)) and DATEDIFF(day,t2.timein,#tmp.timein1)=0
and t2.tId>#tmp.tId order by tId asc)
update #tmp
set TimeIn2 = (select TimeIn from tblTime where tId=tId2),
TimeOut2 = (select [TimeOut] from tblTime where tId=tId2)
update #tmp set TotalMins = (
isnull(DATEDIFF(minute,timein1,timeout1),0)+
isnull(DATEDIFF(minute,timein2,timeout2),0)
)
select * from #tmp order by TimeIn1
drop table #tmp