我有以下sql查询。
SELECT E.EMPID, TM.LOGIN, TM.LOGOUT,
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)/60)+':'+
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)%60) AS WORKEDHOURS
FROM EMPLOYEE_DETAILS E
LEFT JOIN TIMING_DETAILS TM
ON(E.EMPID=TM.EMPID)
WHERE TM.LOGIN BETWEEN '2013-09-02' AND DATEADD(DAY, 1, '2013-09-03')
GROUP BY E.EMPID,TM.LOGIN,TM.LOGOUT
ORDER BY E.EMPID
运行上面的查询后，我得到以下输出：
EMPID LOGIN LOGOUT WORKEDHOURS
1 2-9-2013 9:30 2-9-2013 1:00 3:30
1 2-9-2013 2:00 2-9-2013 6:00 4:00
2 2-9-2013 9:30 2-9-2013 2:00 4:30
2 2-9-2013 3:00 2-9-2013 6:00 3:00
1 3-9-2013 9:30 3-9-2013 6:00 8:30
但是，我想显示结果如下，任何人请在这里帮忙，谢谢
EMPID LOGIN LOGOUT WORKEDHOURS
1 2-9-2013 9:30 2-9-2013 1:00 7:30
1 2-9-2013 2:00 2-9-2013 6:00
2 2-9-2013 9:30 2-9-2013 2:00 7:30
2 2-9-2013 3:00 2-9-2013 6:00
1 3-9-2013 9:30 3-9-2013 6:00 8:30
答案 0 :(得分：0)
SELECT T.EMPID, T.LOGIN, T.LOGOUT, X.WORKEDHOURS
FROM TIMING_DETAILS T
INNER JOIN
(SELECT EMPID, DATEPART(YEAR, TM.LOGIN) YEAR, DATEPART(MONTH, TM.LOGIN) MONTH, DATEPART(DAY, TM.LOGIN) DAY,
CAST(SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT))/60 AS VARCHAR(MAX))
+ ':'
+ CAST(SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT))%60 AS VARCHAR(MAX)) AS WORKEDHOURS
FROM TIMING_DETAILS TM
WHERE TM.LOGIN BETWEEN '2013-09-02' AND DATEADD(DAY, 1, '2013-09-03')
GROUP BY EMPID, DATEPART(YEAR, TM.LOGIN), DATEPART(MONTH, TM.LOGIN), DATEPART(DAY, TM.LOGIN)) X
ON X.EMPID = T.EMPID
AND DATEPART(YEAR, T.LOGIN) = X.YEAR
AND DATEPART(MONTH, T.LOGIN) = X.MONTH
AND DATEPART(DAY, T.LOGIN) = X.DAY
ORDER BY X.EMPID, X.YEAR DESC, X.MONTH DESC, X.DAY DESC
答案 1 :(得分：0)
SELECT E.EMPID, TM.LOGIN, TM.LOGOUT,WORKHOURS_TABLE.WORKEDHOURS
FROM EMPLOYEE_DETAILS E, TIMING_DETAILS TM ,
(SELECT EMPID,
concat(
sum(hour(timeDIFF(TM.LOGOUT, TM.LOGIN )))
+
sum(minute(timeDIFF(TM.LOGOUT, TM.LOGIN ))) div 60 ,
":" ,
sum(minute(timeDIFF(TM.LOGOUT, TM.LOGIN ))) % 60
)
AS WORKEDHOURS
FROM TIMING_DETAILS TM
GROUP BY EMPID) AS WORKHOURS_TABLE
WHERE E.EMPID = WORKHOURS_TABLE.EMPID AND E.EMPID=TM.EMPID
ORDER BY E.EMPID
输出可能是：
EMPID LOGIN LOGOUT WORKEDHOURS
1 2-9-2013 9:30 2-9-2013 1:00 7:30
1 2-9-2013 2:00 2-9-2013 6:00 7:30
2 2-9-2013 9:30 2-9-2013 2:00 7:30
2 2-9-2013 3:00 2-9-2013 6:00 7:30
如果您想按日期分组，则必须为date和time分别保留一列，然后您可以按日期分组。
我在我的MySQL数据库中尝试了查询，它运行正常。你试试看。
答案 2 :(得分：0)
试试这个 -
SELECT
t.EMPID
, t.LOGIN
, t.LOGOUT
, WORKEDHOURS = CASE WHEN rn = 1 THEN t.WORKEDHOURS END
FROM (
SELECT
TM2.EMPID
, TM2.LOGIN
, TM2.LOGOUT
, TM.WORKEDHOURS
, rn = ROW_NUMBER() OVER (PARTITION BY TM2.EMPID ORDER BY TM2.EMPID)
FROM dbo.TIMING_DETAILS TM2
JOIN (
SELECT
EMPID
, WORKEDHOURS = CAST(SUM(DATEDIFF(MINUTE, [login], LOGOUT)) / 60 AS VARCHAR(100)) + ':' + CAST(SUM(DATEDIFF(MINUTE, [login], LOGOUT)) % 60 AS VARCHAR(10))
FROM dbo.TIMING_DETAILS
WHERE [login] BETWEEN '20130902' AND '20130904'
GROUP BY EMPID
) TM ON TM2.EMPID = TM.EMPID
WHERE EXISTS( -- maybe this is unnecessary
SELECT 1
FROM dbo.EMPLOYEE_DETAILS E
WHERE E.EMPID = TM2.EMPID
)
) t
答案 3 :(得分：0)
在我们的对话之后，以及与其他参与者一起审核您的对话时，我发现您的问题稍微受到限制。你真正想要的输出是这样的：
EMPID DAY WORKEDHOURS
1 2-9-2013 7:30
1 3-9-2013 8:30
2 2-9-2013 7:30
案例1：范围内的单日
如果您只对一天感兴趣，那么您只需在结果中插入日期即可。请尝试以下方法：
SELECT E.EMPID,
'2013-09-02' AS DAY,
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)/60)+':'+
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)%60) AS WORKEDHOURS
FROM EMPLOYEE_DETAILS E
LEFT JOIN TIMING_DETAILS TM
ON (E.EMPID=TM.EMPID)
WHERE TM.LOGIN BETWEEN '2013-09-02' AND DATEADD(DAY, 1, '2013-09-03')
GROUP BY E.EMPID
ORDER BY E.EMPID
您还可以重写查询以使用命名参数作为相关日期。这是一个示例，但要使用它，您必须将值绑定到参数TheDay：
SELECT E.EMPID,
@TheDay AS DAY,
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)/60)+':'+
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)%60) AS WORKEDHOURS
FROM EMPLOYEE_DETAILS E
LEFT JOIN TIMING_DETAILS TM
ON (E.EMPID=TM.EMPID)
WHERE TM.LOGIN BETWEEN @TheDay AND DATEADD(DAY, 1, @TheDay)
GROUP BY E.EMPID
ORDER BY E.EMPID
案例2：范围内的多天
如果每个EMPID需要多行，但每天只需要一行，那么事情会更复杂。我认为有可能在几个（也许是所有）DBMS中，但可能需要使用非便携式功能。我们需要做的是GROUP BY LOGIN日期时间值的日期部分。
如果您使用的是SQL Server 2008（或更新版本），则可以尝试以下操作：
SELECT E.EMPID,
CAST(TM.LOGIN AS DATE) AS DAY,
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)/60)+':'+
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)%60) AS WORKEDHOURS
FROM EMPLOYEE_DETAILS E
LEFT JOIN TIMING_DETAILS TM
ON (E.EMPID=TM.EMPID)
WHERE TM.LOGIN BETWEEN '2013-09-02' AND DATEADD(DAY, 1, '2013-09-03')
GROUP BY E.EMPID, CAST(TM.LOGIN AS DATE)
ORDER BY E.EMPID
请注意，我无法测试此查询。例如，您可能必须GROUP BY DAY而不是重复表达。
案例2B：范围内的多天，首次登录和最后一次登录
您是否希望每个EMPID每天一行，但第一次登录时间和上次注销时间？您可以尝试以下方法：
SELECT E.EMPID,
MIN(TM.LOGIN) AS FIRSTLOGIN,
MAX(TM.LOGOUT) AS LASTLOGOUT,
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)/60)+':'+
SUM(DATEDIFF(MINUTE,TM.LOGIN,TM.LOGOUT)%60) AS WORKEDHOURS
FROM EMPLOYEE_DETAILS E
LEFT JOIN TIMING_DETAILS TM
ON (E.EMPID=TM.EMPID)
WHERE TM.LOGIN BETWEEN '2013-09-02' AND DATEADD(DAY, 1, '2013-09-03')
GROUP BY E.EMPID, CAST(TM.LOGIN AS DATE)
ORDER BY E.EMPID
同样，我还没有能力测试这个查询。
TESTED CASE：对Visual Studio 2010中本地SQL Server Express数据库的查询
我今天无法在Sql Server上进行测试，但这是我在本地Sql Server Express数据库上在Visual Studio 2010下测试的查询版本。
SELECT
E.EMPID, E.NAME,
MIN(TM.LOGIN) AS FirstLogIn,
MAX(TM.LOGOUT) AS LastLogOut,
CAST(SUM(DATEDIFF(MINUTE, TM.LOGIN, TM.LOGOUT) / 60) AS NVARCHAR)+':'+CAST(SUM(DATEDIFF(MINUTE, TM.LOGIN, TM.LOGOUT) % 60) AS NVARCHAR) AS WorkedHours
FROM
EMPLOYEE_DETAILS AS E LEFT OUTER JOIN TIMING_DETAILS AS TM ON E.EMPID = TM.EMPID
WHERE
TM.LOGIN BETWEEN '2013-09-21' AND DATEADD(DAY, 1, '2013-09-22')
GROUP BY
E.EMPID, E.NAME,
DATEADD(DAY, DATEDIFF(DAY, 0, TM.LOGIN), 0)
ORDER BY
E.EMPID, FirstLogin
测试数据EMPLOYEE_DETAILS：
EMPID NAME
1 Joe
2 Jane
3 Martha
测试数据TIMING_DETAILS：
EMPID LOGIN LOGOUT
1 '2013-09-21 09:00:00' '2013-09-21 10:30:00'
1 '2013-09-21 14:00:00' '2013-09-21 16:00:00'
1 '2013-09-22 08:30:00' '2013-09-22 11:00:00'
2 '2013-09-21 10:45:00' '2013-09-21 13:30:00'
结果：
EMPID NAME FirstLogIn LastLogOut WorkedHours
1 Joe '2013-09-21 09:00:00' '2013-09-21 16:00:00' 3:30
1 Joe '2013-09-22 08:30:00' '2013-09-22 11:00:00' 2:30
2 Jane '2013-09-21 10:45:00' '2013-09-21 13:30:00' 2:45
这对我来说似乎是对的。我的SQL Server Express查询的唯一区别是它不提供DATE类型，因此我不得不修改CAST。哦，我还必须明确地向NVARCHAR投入WorkedHours。
请根据您的环境进行相应调整。
希望这有帮助。