如何通过 SQL 根据某些条件过滤唯一记录

Question

我有一个 SQL 服务器数据库，其中包含一个从考勤设备获取数据的表。

它看起来像这样：

ID	用户身份	日期	时间	指示方向
1988781	25000	2022/01/11	10:02	1
1988782	25000	2022/01/11	10:03	1
1988783	25000	2022/01/11	10:04	1
1988784	25000	2022/01/11	12:30	2
1988785	25000	2022/01/11	12:31	2
1988786	25001	2022/01/11	10:00	1
1988787	25001	2022/01/11	12:30	2
1988788	25002	2022/01/11	10:15	1
1988789	25002	2022/01/11	10:16	1
1988790	25002	2022/01/11	12:19	2

如何过滤每个用户 ID 的数据以使 SignDirection = 1 的时间更短，而 SignDirection = 2 的时间更长？

像是：

ID	用户身份	日期	时间	指示方向
1988781	25000	2022/01/11	10:02	1
1988785	25000	2022/01/11	12:31	2
1988786	25001	2022/01/11	10:00	1
1988787	25001	2022/01/11	12:30	2
1988788	25002	2022/01/11	10:15	1
1988790	25002	2022/01/11	12:19	2

我努力了

select distinct

但没有成功。

Answer 1

您可以计算 2 row_number，向上和向下。
然后过滤它们。

SELECT ID, UserID, [Date], [Time], SignDirection
(
    SELECT *
    , ROW_NUMBER() OVER (PARTITION BY UserID, SignDirection, [Date] ORDER BY [Time] ASC) rn_asc
    , ROW_NUMBER() OVER (PARTITION BY UserID, SignDirection, [Date] ORDER BY [Time] DESC) rn_desc
    FROM your_time_and_attendance_table t
) q
WHERE ((SignDirection = 1 AND rn_asc = 1) OR
       (SignDirection = 2 AND rn_desc = 1)) 
ORDER BY ID, UserID, [Date], [Time], SignDirection

Answer 2

@LukStorms 给出的答案很好，但它需要对第二个行号进行另一种排序。

您可以改为使用具有相同排序的LEAD来检测最后一行

SELECT
  ID,
  UserID,
  [Date],
  [Time],
  SignDirection
FROM
(
    SELECT *
    , ROW_NUMBER() OVER (PARTITION BY UserID, SignDirection, [Date] ORDER BY [Time]) rn_asc
    , LEAD(Time) OVER (PARTITION BY UserID, SignDirection, [Date] ORDER BY [Time]) nextTime
    FROM YourTable t
) q
WHERE ((SignDirection = 1 AND rn_asc = 1) OR
       (SignDirection = 2 AND nextTime IS NULL)) 
ORDER BY
  ID,
  UserID,
  [Date],
  [Time],
  SignDirection;

db<>小提琴