Question

我写了这个查询，根据他的INTime和OutTime显示员工的工作时间，计算它之间的差异，并计算2班的OVETIME如果时间超过s002，则18:00 pm仅计算加班时间;对于班次s001，它仅计算加班17:00。

这就是全部。

现在的问题是，如果某人在某一天没有来，那么我应该将ABSENT放在Total WorkingTime和Overtime中，或者如果仅提及他的In Time，我应该将Not Defined放入WORKING TIME和OVER TIME。

查询：

with times as (
SELECT    t1.EmplID
        , t3.EmplName
        , min(t1.RecTime) AS InTime
        , max(t2.RecTime) AS [TimeOut]
        , t4.ShiftId as ShiftID
        , t4.StAtdTime as ShStartTime
        , t4.EndAtdTime as ShEndTime
        , cast(min(t1.RecTime) as datetime) AS InTimeSub
        , cast(max(t2.RecTime) as datetime) AS TimeOutSub
        , t1.RecDate AS [DateVisited]
FROM  AtdRecord t1 
INNER JOIN 
      AtdRecord t2 
ON    t1.EmplID = t2.EmplID 
AND   t1.RecDate = t2.RecDate
AND   t1.RecTime < t2.RecTime
inner join 
      HrEmployee t3 
ON    t3.EmplID = t1.EmplID 
inner join AtdShiftSect t4
ON t3.ShiftId = t4.ShiftId
group by 
          t1.EmplID
        , t3.EmplName
        , t1.RecDate
        , t4.ShiftId 
        , t4.StAtdTime 
        , t4.EndAtdTime
)
SELECT 
 EmplID
,EmplName
,ShiftId As ShiftID
,InTime
,[TimeOut]
,convert(char(5),cast([TimeOutSub] - InTimeSub as time), 108) TotalWorkingTime
,[DateVisited]
,CONVERT(char(5),CASE WHEN  CAST([TimeOutSub] AS DATETIME) >= ShEndTime And ShiftID = 'S002'
     Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST(ShEndTime AS DATETIME),  CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
     WHEN  CAST([TimeOutSub] AS DATETIME) >= ShEndTime And ShiftID = 'S001'
     Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST(ShEndTime AS DATETIME),  CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
     ELSE '00:00' END, 108) AS OverTime
FROM times order by EmplID, ShiftID, DateVisited

Answer 1

试试这个

查找DateVisited NULL或NOt并进一步移动

with times as (
SELECT    t1.EmplID
        , t3.EmplName
        , min(t1.RecTime) AS InTime
        , max(t2.RecTime) AS [TimeOut]
        , t4.ShiftId as ShiftID
        , t4.StAtdTime as ShStartTime
        , t4.EndAtdTime as ShEndTime
        , cast(min(t1.RecTime) as datetime) AS InTimeSub
        , cast(max(t2.RecTime) as datetime) AS TimeOutSub
        , t1.RecDate AS [DateVisited]
FROM  AtdRecord t1 
INNER JOIN 
      AtdRecord t2 
ON    t1.EmplID = t2.EmplID 
AND   t1.RecDate = t2.RecDate
AND   t1.RecTime < t2.RecTime
inner join 
      HrEmployee t3 
ON    t3.EmplID = t1.EmplID 
inner join AtdShiftSect t4
ON t3.ShiftId = t4.ShiftId
group by 
          t1.EmplID
        , t3.EmplName
        , t1.RecDate
        , t4.ShiftId 
        , t4.StAtdTime 
        , t4.EndAtdTime
)
SELECT 
 EmplID
,EmplName
,ShiftId As ShiftID
,InTime
,[TimeOut]
,convert(char(5),cast([TimeOutSub] - InTimeSub as time), 108) TotalWorkingTime
,[DateVisited]
,CASE WHEN [InTime] IS NOT NULL AND [TimeOut] IS NOT NULL THEN
     CONVERT(char(5),CASE WHEN  CAST([TimeOutSub] AS DATETIME) >= ShEndTime And ShiftID = 'S002' Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST(ShEndTime AS DATETIME),CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
                          WHEN  CAST([TimeOutSub] AS DATETIME) >= ShEndTime And ShiftID = 'S001' Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST(ShEndTime AS DATETIME),  CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
      ELSE '00:00' END, 108) 
 ELSE 'ABSENT' END AS OverTime
 FROM times order by EmplID, ShiftID, DateVisited

把缺席放在列中

1 个答案: