显示特定人的结果

Question

我已经把这个查询写成显示加班，如果有人工作时间超过下午18:00时钟时间，无论他/她来的时间或他/她工作的工作时间，这都超出了主题。它只显示OVERTIME，如果有人在18:00 pm以上工作，那就是它，例如InTime = 15:16 pm，TimeOut = 18:31然后Overtime = 00:31。所以我写了下面给出的这个问题，但现在问题是下午17:00离职的人数很少，他们的超时时间应该在下午17:00之后计算（他在上午17:00以后工作了多少分钟+小时），所以那些具体的员工超时应该在17:00之后计算

那些员工是：

EmplID 5, 43, 11

所以只有这些加班才能在17:00之后计算出来。

查询：

with times as (
SELECT    t1.EmplID
        , t3.EmplName
        , min(t1.RecTime) AS InTime
        , max(t2.RecTime) AS [TimeOut]
        , 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 
group by 
          t1.EmplID
        , t3.EmplName
        , t1.RecDate
)
SELECT EmplID
,EmplName
,InTime
,[TimeOut]
,[DateVisited]
,convert(char(5),cast([TimeOutSub] - InTimeSub as time), 108) totaltime
,CONVERT(char(5),CASE WHEN  CAST([TimeOutSub] AS DATETIME) >= '18:00' Then     
 LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST('18:00' AS DATETIME),   
 CAST([TimeOutSub] AS DATETIME)),0), 108),5) ELSE '00:00' END, 108) AS OVERTIME
FROM times

输出：

Answer 1

尝试更换：

CASE WHEN  CAST([TimeOutSub] AS DATETIME) >= '18:00' Then     
 LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST('18:00' AS DATETIME),   
 CAST([TimeOutSub] AS DATETIME)),0), 108),5) ELSE '00:00' END, 108) AS OVERTIME

用

CASE WHEN  CAST([TimeOutSub] AS DATETIME) >= '18:00' And EmplId Not In (5,43,11)
     Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST('18:00' AS DATETIME),  CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
     WHEN  CAST([TimeOutSub] AS DATETIME) >= '17:00' And EmplId In (5,43,11)
     Then  LEFT(CONVERT(varchar(12), DATEADD(ms, DATEDIFF(ms, CAST('17:00' AS DATETIME),  CAST([TimeOutSub] AS DATETIME)),0), 108),5) 
     ELSE '00:00' END, 108) AS OVERTIME

基本上，我只在WHEN部分添加了多个条件。 When time > 18:00和emplid not IN (5,43,11)以及time > 17:00 and emplid in (5,43,11).

的另一个条件

将其理解为伪代码可能更容易：

CASE WHEN  TIME >= '18:00' And EmplId Not In (5,43,11)
     Then  -- Calculate overtime at 18:00
     WHEN  TIME >= '17:00' And EmplId In (5,43,11)
     Then  -- Calculate overtime at 17:00
     ELSE '00:00' 
     END OVERTIME