Question

我的表格如下：

+-----+---------------------+-----+-----+
| id  | date                | uid | sid |
+-----+---------------------+-----+-----+
| 454 | 2018-11-18 10:12:16 | 206 |  10 | 
| 456 | 2018-11-18 10:53:37 | 206 |  20 | O
| 467 | 2018-11-18 13:00:02 | 206 |  10 | C
| 469 | 2018-11-18 14:50:33 | 206 |  10 | 
| 452 | 2018-11-18 07:11:56 | 208 |  10 | 
| 470 | 2018-11-18 15:01:38 | 208 |  20 | 
| 455 | 2018-11-18 10:51:29 | 209 |  10 | 
| 458 | 2018-11-18 11:30:45 | 209 |  20 | O
| 459 | 2018-11-18 11:35:08 | 209 |  20 | 
| 460 | 2018-11-18 11:48:24 | 209 |  20 | 
| 462 | 2018-11-18 11:55:12 | 209 |  20 | 
| 464 | 2018-11-18 12:09:20 | 209 |  10 | C
| 465 | 2018-11-18 12:30:15 | 209 |  10 | 
| 468 | 2018-11-18 14:00:58 | 209 |  10 | 
| 471 | 2018-11-18 17:25:19 | 209 |  20 | O
| 472 | 2018-11-18 18:52:24 | 209 |  10 | C
| 453 | 2018-11-18 08:38:23 | 212 |  10 | 
| 457 | 2018-11-18 11:29:03 | 212 |  20 | O
| 461 | 2018-11-18 11:49:54 | 212 |  20 | 
| 463 | 2018-11-18 12:08:49 | 212 |  10 | C
| 466 | 2018-11-18 12:52:11 | 212 |  10 | 
+-----+---------------------+-----+-----+

我需要获得MIN（打开）和MIN（关闭）状态当用户打开sid = 20且关闭时间sid = 10时而且我每个用户有多个会话

当我使用[Retrieving the last record in each group - MySQL]的报价

select 
t2.id, t2.date, t2.uid, t2.sid
from (
select t.*,
@r:= case when @g = t.uid
     then
          case when @sg = t.sid
          then 
            case when @sr <> t.date
            then  @r+1 
            else @r end 
          else 1 end 
      else 1 end r,
@g:= uid g,
@sg:= t.sid sg,
@sr:= t.date sr
from messages t
cross join (select @g:=null,@sg:=null,@r:=null) t1
order by t.uid,t.sid,t.date
  ) t2
where t2.r = 1 and sid = 20 or t2.r = 2 and sid = 10
order by uid,date

我得到

+-----+---------------------+-----+----+
| id  | date                | uid |sid |
+-----+---------------------+-----+----+
| 456 | 2018-11-18 10:53:37 | 206 | 20 |
| 467 | 2018-11-18 13:00:02 | 206 | 10 |
| 470 | 2018-11-18 15:01:38 | 208 | 20 |
| 458 | 2018-11-18 11:30:45 | 209 | 20 |
| 464 | 2018-11-18 12:09:20 | 209 | 10 |
| 457 | 2018-11-18 11:29:03 | 212 | 20 |
| 463 | 2018-11-18 12:08:49 | 212 | 10 |
+-----+---------------------+-----+----+

问题是我怎么能得到这个？

+-----+---------------------+-----+----+
| id  | date                | uid |sid |
+-----+---------------------+-----+----+
| 456 | 2018-11-18 10:53:37 | 206 | 20 |
| 467 | 2018-11-18 13:00:02 | 206 | 10 |
| 458 | 2018-11-18 11:30:45 | 209 | 20 |
| 464 | 2018-11-18 12:09:20 | 209 | 10 |
| 471 | 2018-11-18 17:25:19 | 209 | 20 |
| 472 | 2018-11-18 19:05:38 | 209 | 10 |
| 457 | 2018-11-18 11:29:03 | 212 | 20 |
| 463 | 2018-11-18 12:08:49 | 212 | 10 |
+-----+---------------------+-----+----+

Answer 1

假设您的id值随着date值的增加而增加（或者如果您不太在意id的值），并且您不介意公开并在同一行关闭时间，此查询将为您提供所需的结果，而无需使用变量can be unreliable：

SELECT MIN(m1.id) AS m1_id
     , MIN(m1.uid) AS uid
     , MIN(m1.sid) AS sid
     , MIN(m1.date) AS m1_date
     , MIN(m2.id) AS m2_id
     , MIN(m2.sid) AS m2_sid
     , m2.date AS m2_date
FROM messages m1
LEFT JOIN messages m2 ON m2.uid = m1.uid AND m2.sid != m1.sid AND m2.sid = 10 AND
    m2.date = (SELECT MIN(date)
               FROM messages m3 
               WHERE m3.uid = m2.uid AND m3.sid = 10 AND m3.date > m1.date)
WHERE m2.id IS NOT NULL
GROUP BY m2.date

输出：

m1_id   uid     sid     m1_date                 m2_id   m2_sid  m2_date
456     206     20      2018-11-18 10:53:37     467     10      2018-11-18 13:00:02
458     209     20      2018-11-18 11:30:45     464     10      2018-11-18 12:09:20
471     209     20      2018-11-18 17:25:19     472     10      2018-11-19 18:52:24
457     212     20      2018-11-18 11:29:03     463     10      2018-11-18 12:08:49

Demo on dbfiddle

MIN（）DateTime记录检索-MySQL

1 个答案: