Question

我已根据Intime和outTime编写查询来计算员工的总工作时间和加班时间，但无法计算出另一列“IsAbsent”。例如如果某人没有任何具体日期，他的记录将不会出现在该特定日期，那么新的列IsAbsent应该包含否则存在。

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 TimeOutSub - InTimeSub >= '08:01' then 
cast(TimeOutSub - dateadd(hour, 8, InTimeSub) as time) else '00:00' end, 108) as overtime
FROM times

enter image description here

Answer 1

我在阅读你的桌子时有点麻烦...但我们假设我们有两张桌子：员工和时间......你需要做一个左连接：

 select e.*, t.* 
 from employee e 
     left join time t on e.employee_id = t.employee_id

当t.employee_id为null时，您将没有时间记录......所以让我们构建它：

 select e.*
      IsAbsent = case 
                     when t.employee_id is null then 'ABSENT' 
                     else 'PRESENT' 
                 end 
 from employee e 
     left join time t on e.employee_id = t.employee_id

下一步可能会对时间表（例如2014年的所有内容）施加约束，所以让我们在下面构建：

 select e.*,
      IsAbsent = case 
                     when t.employee_id is null then 'ABSENT' 
                     else 'PRESENT' 
                 end 
 from employee e 
     left join time t on e.employee_id = t.employee_id and 
                datepart(y, t.timestamp) = 2014

请注意我已将约束放在联接中，这是至关重要的。

这有意义吗？

把新栏目记录下来

1 个答案: