如何使用sql server 2008获取特定月份员工的总工作时间? [英] How can I get the total worked hours by employee in a specific month using sql server 2008?
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问题描述
我正在尝试显示每个员工在特定月份的总工作时间,我正在使用一个表Emp_Attendance,其中包含(EmpID,_date Times_In,Times_Out)
我尝试这个sql查询,但它给了我一个错误的结果(正确的结果是14:00,但它给了我104:16)
I'm trying to display the total worked hours for each employee in a specific month, I'm using a table "Emp_Attendance" which contain the (EmpID,_date Times_In, Times_Out)
I try this sql query but it gives me a wrong result (the right result is 14:00 but it gives me 104:16)
WITH att As
( SELECT EmpID ,
Times_In, Times_Out,
ROW_NUMBER() OVER (PARTITION BY EmpID ORDER BY Times_In) As RNI,
ROW_NUMBER() OVER (PARTITION BY EmpID ORDER BY Times_Out DESC) As RNO
FROM dbo.Emp_Attendance
WHERE
Month(Times_Out) = 6 )
SELECT EmpID,
CONVERT(varchar(3),(Datediff (MINUTE, Max(CASE WHEN RNI = 1 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 4 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 2 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 3 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 3 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 2 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 4 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 1 THEN Times_Out ELSE Null END)) ) /60 )
+':' + RIGHT('0' + CONVERT(varchar(2),
(Datediff (MINUTE, Max(CASE WHEN RNI = 1 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 4 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 2 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 3 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 3 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 2 THEN Times_Out ELSE Null END))+
Datediff (MINUTE, Max(CASE WHEN RNI = 4 THEN Times_In ELSE Null END), Max(CASE WHEN RNO = 1 THEN Times_Out ELSE Null END))) %60 ) ,2)
as total
FROM att
GROUP BY EmpID
推荐答案
目前还不完全清楚为什么要检查行号以及为什么你有一个CTE。
从您描述的问题来看,这个简单的查询将为您提供所需的结果
It's not entirely clear why you are checking Row Numbers nor why you have a CTE.
From the problem as you have described it, this simple query will give you the results you need
select EmpId, SUM(DATEDIFF(MINUTE, Times_In, Times_Out)) / 60.0
from Emp_Attendance
where MONTH(Times_In) = 6
GROUP BY EmpId
请注意该部门的 .0
,以确保我们不会失去任何部分时间。
Note the .0
on the division to ensure we don't lose any partial hours.
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