出勤时间和超时SQL查询

Question

我想在下面的数据得到帮助，我们有手冲床节省出入境记录，各出入境记录可以存储不同行的雇员ID

EMP ID | EntryTime    | ExitTime 
11769  2015-02-01 08:00:00  NULL 
11769  NULL      2015-02-01 13:00:00 

基本上每两个入口和出。

我想获得的时间和时间出去一天。正如你所看到的，这是一天中的时间和时间，但它存储了两条记录。 所以在这方面我很困惑如何解决这个问题。还要计算工时

一个帮助将是非常赞赏

Answer 1

在这种情况下使用子查询。

SELECT at1.empId ,at1.EntryTime , 
     (SELECT MIN(a2.ExitTime) 
      FROM  attendance at2 
      WHERE  at1.empId = at2.empId 
        AND at2.ExitTime >at1.EntryTime) ExitTime 
FROM attendance at1 
WHERE at1.EntryTime IS NOT NULL 

Answer 2

如果你的结构正是你所提供的样品中，和你没有背靠背挥笔（如员工可能刷他的卡两次想第一个未提交），那么你可以做这样的查询：

WITH InOut (empId, EntryTime, ExitTime) as 
(SELECT empId , 
     a1.EntryTime , 
     (SELECT MIN(a2.ExitTime) 
      FROM  attendance a2 
      WHERE  a1.empId = a2.empId 
        AND a1.EntryTime < a2.ExitTime 
     ) 
FROM attendance a1 
WHERE EntryTime IS NOT NULL 
) 
SELECT empId , 
     EntryTime , 
     ExitTime, 
     DATEDIFF(Minute, EntryTime, ExitTime) AS minutes FROM InOut; 

编辑： 我不明白您的评论。下面是一个完整的示例：

DECLARE @attendance TABLE 
    (
     EMPID INT , 
     EntryTime DATETIME NULL , 
     ExitTime DATETIME NULL 
    ); 

INSERT @attendance 
     (EMPID, EntryTime, ExitTime) 
VALUES (11769, '2015-02-01 08:00:00', NULL), 
     (11769, NULL, '2015-02-01 13:00:00'), 
     (11769, '2015-02-02 09:00:00', NULL), 
     (11769, NULL, '2015-02-02 12:00:00'), 
     (11769, '2015-02-03 08:00:00', NULL), 
     (11769, NULL, '2015-02-03 13:00:00'), 
     (1, '2015-02-01 08:10:00', NULL), 
     (1, NULL, '2015-02-01 13:10:00'), 
     (1, '2015-02-02 09:10:00', NULL), 
     (1, NULL, '2015-02-02 12:10:00'), 
     (1, '2015-02-03 08:10:00', NULL), 
     (1, NULL, '2015-02-03 13:10:00'), 
     (2, '2015-02-01 08:30:00', NULL), 
     (2, NULL, '2015-02-01 13:30:00'), 
     (2, '2015-02-02 09:30:00', NULL), 
     (2, NULL, '2015-02-02 12:30:00'), 
     (2, '2015-02-03 08:30:00', NULL); 

WITH InOut (empId, EntryTime, ExitTime) 
      AS (SELECT EMPID , 
         a1.EntryTime , 
         (SELECT MIN(a2.ExitTime) 
          FROM  @attendance a2 
          WHERE  a1.EMPID = a2.EMPID 
            AND a1.EntryTime < a2.ExitTime 
         ) 
       FROM  @attendance a1 
       WHERE EntryTime IS NOT NULL 
      ) 
    SELECT empId , 
      EntryTime , 
      ExitTime , 
      DATEDIFF(MINUTE, EntryTime, ExitTime) AS minutes 
    FROM InOut; 

，其结果是：

empId EntryTime    ExitTime    minutes 
11769 2015-02-01 08:00:00.000 2015-02-01 13:00:00.000 300 
11769 2015-02-02 09:00:00.000 2015-02-02 12:00:00.000 180 
11769 2015-02-03 08:00:00.000 2015-02-03 13:00:00.000 300 
1  2015-02-01 08:10:00.000 2015-02-01 13:10:00.000 300 
1  2015-02-02 09:10:00.000 2015-02-02 12:10:00.000 180 
1  2015-02-03 08:10:00.000 2015-02-03 13:10:00.000 300 
2  2015-02-01 08:30:00.000 2015-02-01 13:30:00.000 300 
2  2015-02-02 09:30:00.000 2015-02-02 12:30:00.000 180 
2  2015-02-03 08:30:00.000 NULL     NULL