sql server 获取最小时间？

Question

问题来了，

我有一个名为 attendance 的表，有 4 列：

[Username] varchar(256)
[Date] varchar(256)
[Time] varchar(256)
[Action] varchar(256)

我希望实现的是获得第一次入住和退房时间和最后一次入住和退房时间。如果表中只有一个签入和签出行，则返回相同。

这样的示例数据：

Username | Date       | Time     | Action
--------------------------------------------
User1    | 01/12/2012 | 12:54:41 | Check In
User1    | 01/12/2012 | 18:26:36 | Check Out
User1    | 01/12/2012 | 18:44:17 | Check In
User1    | 01/12/2012 | 22:05:31 | Check Out

我的预期结果如下所示：

Output:
Username | Date       | First In | First Out | Last In  | Last Out
User1    | 01/12/2012 | 12:54:41 | 18:26:36  | 18:44:17 | 22:05:31

我尝试了这条 SQL 语句：

SELECT [USERNAME], [DATE]
, min(case when [action] = 'Clock In' then [time] else '' end) as 'First in'
, min(case when [action] = 'Clock Out' then [time] else '' end )as 'First out'
, max(case when [action] = 'Clock In' then [time] else '' end) as 'Last in'
, max(case when [action] = 'Clock Out' then [time] else '' end )as 'Last out'
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012'
group by [username],[date]

它返回意外的结果，即我的最小值总是 '' 或为空，但我的最大值都是 “后进”和“后出”的值正确。

Output:
Username | Date       | First In | First Out | Last In  | Last Out
User1    | 01/12/2012 |          |           | 18:44:17 | 22:05:31

我的 SQL 有什么问题吗？

谁能给点建议？

Answer 1

我会使用 row_number() 来做这件事略有不同：

select username, date,
  max(case when [action] = 'Check In' and rn=1 then time end) FirstIn,
  max(case when [action] = 'Check Out' and rn=1 then time end) FirstOut,
  max(case when [action] = 'Check In' and rn=2 then time end) LastIn,
  max(case when [action] = 'Check Out' and rn=2 then time end) LastOut
from
(
  select username, cast(date as date) date, 
    cast(time as time) time, 
    action,
    row_number() 
      over(partition by username, cast(date as date), action
           order by cast(time as time)) rn
  from attendance
) src
group by username, date;

见SQL Fiddle with Demo

结果是：

| USERNAME |       DATE |          FIRSTIN |         FIRSTOUT |           LASTIN |          LASTOUT |
-----------------------------------------------------------------------------------------------------
|    User1 | 2012-01-12 | 12:54:41.0000000 | 18:26:36.0000000 | 18:44:17.0000000 | 22:05:31.0000000 |

我强烈建议您将date 和time 存储为datetime 数据类型。

注意：我将值从 Clock In/Clock Out 更改为 Check In/Check Out 以匹配您提供的示例数据。

Answer 2

试试这个：

SELECT [USERNAME], [DATE]
, min(case when [action] = 'Clock In' then [time]  end) as 'First in'
, min(case when [action] = 'Clock Out' then [time] end )as 'First out'
, max(case when [action] = 'Clock In' then [time] end) as 'Last in'
, max(case when [action] = 'Clock Out' then [time] end )as 'Last out'
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012'
group by [username],[date]

你知道问题出在哪里吗？

Answer 3

我的解决方案是创建一个存储过程：

DECLARE @first_in VARCHAR, @first_out VARCHAR, @last_in VARCHAR, @last_out VARCHAR
BEGIN
SELECT TOP 1
     @first_in = time
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012' AND [ACTION] = 'first in'
ORDER BY time ASC


SELECT TOP 1
     @first_in = time
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012' AND [ACTION] = 'first out'
ORDER BY time ASC

SELECT TOP 1
     @last_in = time
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012' AND [ACTION] = 'first in'
ORDER BY time DESC

SELECT TOP 1
     @last_out = time
FROM attendance
WHERE [USERNAME] = 'User1' AND [DATE] = '01/12/2012' AND [ACTION] = 'first out'
ORDER BY time DESC

SELECT 'User1','01/12/2012', @first_in, @first_out, @last_in, @last_out
END