SQL考勤查询

Question

最近我从 MS Access 切换到 SQL 服务器。由于这个开关，我在使一个 SQL 查询工作时遇到了问题。

这就是当前 table 在 SQL 中的样子。

这是我试图从查询中得到的结果：

以前我可以通过以下查询使其在 MS Access 中工作：

SELECT m.UserEnrollNumber, m.Checktime AS TimeIn, (SELECT Min(s.Checktime)
    FROM CheckInOut1 s
    WHERE s.UserEnrollNumber = m.UserEnrollNumber
    AND s.Checktime > m.Checktime
    AND s.Checktime <= Int(m.Checktime) + 1) AS TimeOut
FROM CheckInOut1 AS m
WHERE ((((SELECT COUNT(*) 
    FROM CheckInOut1 s 
    WHERE s.UserEnrollNumber = m.UserEnrollNumber
    AND s.Checktime <= m.Checktime
    AND s.Checktime >= INT(m.Checktime)) Mod 2)=1));

以下查询作为@GMB 的回答：

select
        employee_id,
        min(time_in_out) check_in,
        max(time_in_out) check_out
    from (
        select t.*, row_number() over(partition by employee_id order by time_in_out) - 1 rn
        from mytable t
    ) t
    group by employee_id, floor(rn / 2) 
    order by employee_id, floor(rn / 2)

来自 SQL table:

给我以下结果：

似乎显示了最小行和最大行，但中间的行没有显示。

来自@Gordon Linoff 的以下查询：

SELECT cio.EmployeeID, cio.TimeInOut AS CheckIn, 
       cio.TimeInOut as CheckOut
FROM (SELECT cio.*,
             ROW_NUMBER() OVER (PARTITION BY cio.EmployeeID, CONVERT(date, cio.TimeInOut) ORDER BY cio.TimeInOut) as seqnum,
             LEAD(cio.TimeInOut) OVER (PARTITION BY cio.EmployeeID, CONVERT(date, cio.TimeInOut) ORDER BY cio.TimeInOut) as next_TimeInOut
      FROM CheckInOut22 cio
     ) cio
WHERE seqnum % 2 = 1;

给我以下结果：

签入与签出相同。

所有帮助将不胜感激。

Answer 1

这在 SQL 服务器中简单得多。使用 window 函数：

SELECT cio.EmployeeID, cio.TimeInOut AS CheckIn, 
       cio.next_TimeInOut as CheckOut
FROM (SELECT cio.*,
             ROW_NUMBER() OVER (PARTITION BY cio.EmployeeID, CONVERT(date, cio.TimeInOut) ORDER BY cio.TimeInOut) as seqnum,
             LEAD(cio.TimeInOut) OVER (PARTITION BY cio.EmployeeID, CONVERT(date, cio.TimeInOut) ORDER BY cio.TimeInOut) as next_TimeInOut
      FROM CheckInOut cio
     ) cio
WHERE seqnum % 2 = 1;

SQL考勤查询

SQL Time Attendance Query

sql

sql-server

time-and-attendance