Question

如何输出样本记录，看起来像预期结果？

示例数据库

 Table Sample1
 Column Status      Int
 Column dtDateTime  dateTime
 Column User        varchar(50)

记录

 User     dtDateTime                   Status
 User1    2017-05-19 08:00:00.000      1
 User1    2017-05-19 17:05:00.000      2
 User1    2017-05-20 09:05:00.000      1
 User1    2017-05-21 17:35:00.000      2

记录1作为时间输入，记录2作为列状态的超时

预期结果

 User         WORK-DATE      Time-IN         Time-OUT       Status
 user1        2017-05-19     08:00:00.000    17:05:00.000   COMPLETE
 User1        2017-05-20     09:05:00.000    NULL           NO OUT
 User1        2017-05-21     NULL            17:35:00.000   NO IN

Answer 1

我的水晶球提出了这个问题：

SELECT   User, 
         CONVERT(DATE, dtDateTime) AS [WORK-DATE],
         CASE WHEN MIN(STATUS) <> MAX(STATUS) THEN 'COMPLETE'
              WHEN MAX(STATUS) = 1 THEN 'NO OUT'
              WHEN MIN(STATUS) = 2 THEN 'NO IN'
              END           
FROM     Sample1
GROUP BY User, CONVERT(DATE, dtDateTime)

编辑：水晶球没有显示早期的TIME栏

SELECT   User, 
         CONVERT(DATE, dtDateTime) AS [WORK-DATE],
         MAX(CASE WHEN STATUS = 1 THEN CONVERT(TIME, dtDateTime) END) AS [TIME-IN],
         MAX(CASE WHEN STATUS = 2 THEN CONVERT(TIME, dtDateTime) END) AS [TIME-OUT],
         CASE WHEN MIN(STATUS) <> MAX(STATUS) THEN 'COMPLETE'
              WHEN MAX(STATUS) = 1 THEN 'NO OUT'
              WHEN MIN(STATUS) = 2 THEN 'NO IN'
              END AS Status           
FROM     Sample1
GROUP BY User, CONVERT(DATE, dtDateTime)

我怎样才能实现这个SQL查询？

1 个答案: