SQL：如何通过两列的唯一组合进行分组？

Question

语境：

• 表message的from_user_id和to_user_id
• 用户应该看到最近显示的最后一条消息的对话
• 会话由多个消息组成，这些消息具有相同的用户ID组合（用户发送消息，用户接收消息）

表内容：

+-------------------------------------------------+--------------+------------+
| text                                            | from_user_id | to_user_id |
+-------------------------------------------------+--------------+------------+
| Hi there!                                       |           13 |         14 | <- Liara to Penelope
| Oh hi, how are you?                             |           14 |         13 | <- Penelope to Liara
| Fine, thanks for asking. How are you?           |           13 |         14 | <- Liara to Penelope
| Could not be better! How are things over there? |           14 |         13 | <- Penelope to Liara
| Hi, I just spoke to Penelope!                   |           13 |         15 | <- Liara to Zara
| Oh you did? How is she?                         |           15 |         13 | <- Zara to Liara
| Liara told me you guys texted, how are things?  |           15 |         14 | <- Zara to Penelope
| Fine, she's good, too                           |           14 |         15 | <- Penelope to Zara
+-------------------------------------------------+--------------+------------+

我的尝试是按from_user_id和to_user_id ，但我显然得到了用户收到的一组消息和用户发送的另一组消息。

SELECT text, from_user_id, to_user_id,created FROM message 
WHERE from_user_id=13 or to_user_id=13
GROUP BY from_user_id, to_user_id
ORDER BY created DESC

得到我：

+-------------------------------+--------------+------------+---------------------+
| text                          | from_user_id | to_user_id | created             |
+-------------------------------+--------------+------------+---------------------+
| Oh you did? How is she?       |           15 |         13 | 2017-09-01 21:45:14 | <- received by Liara
| Hi, I just spoke to Penelope! |           13 |         15 | 2017-09-01 21:44:51 | <- send by Liara
| Oh hi, how are you?           |           14 |         13 | 2017-09-01 17:06:53 |
| Hi there!                     |           13 |         14 | 2017-09-01 17:06:29 |
+-------------------------------+--------------+------------+---------------------+

虽然我想：

+-------------------------------+--------------+------------+---------------------+
| text                          | from_user_id | to_user_id | created             |
+-------------------------------+--------------+------------+---------------------+
| Oh you did? How is she?       |           15 |         13 | 2017-09-01 21:45:14 | <- Last message of conversation with Zara
| Oh hi, how are you?           |           14 |         13 | 2017-09-01 17:06:53 |
+-------------------------------+--------------+------------+---------------------+

我怎样才能做到这一点？

编辑：使用least或greatest也不会导致所需的结果。 它会对条目进行正确分组，但正如您在结果中看到的那样，最后一条消息不正确。

+----+-------------------------------------------------+------+---------------------+--------------+------------+
| id | text                                            | read | created             | from_user_id | to_user_id |
+----+-------------------------------------------------+------+---------------------+--------------+------------+
|  8 | Oh you did? How is she?                         | No   | 2017-09-01 21:45:14 |           15 |         13 |
|  5 | Could not be better! How are things over there? | No   | 2017-09-01 17:07:47 |           14 |         13 |
+----+-------------------------------------------------+------+---------------------+--------------+------------+

Answer 1

与＃13的最后一次对话？ 在更新的DBMS中，您可以使用row_number()来查找这些内容。 在MySQL中，您可以使用not exists ，以确保对话伙伴没有后期帖子。 你可以通过from_user_id + to_user_id - 13顺便找到合作伙伴的号码。 （比较两个记录时，您可以使用from_user_id + to_user_id 。）

select text, from_user_id, to_user_id, created
from message m1
where 13 in (from_user_id, to_user_id)
and not exists
(
  select *
  from message m2
  where 13 in (m2.from_user_id, m2.to_user_id)
  and m2.from_user_id + m2.to_user_id = m1.from_user_id + m1.to_user_id
  and m2.created > m1.created
);

Answer 2

执行所需操作的一种方法是使用相关子查询，以查找匹配对话的最小创建日期/时间：

SELECT m.*
FROM message m
WHERE 13 in (from_user_id, to_user_id) AND
      m.created = (SELECT MAX(m2.created)
                   FROM message m2
                   WHERE (m2.from_user_id = m.from_user_id AND m2.to_user_id = m.to_user_id) OR
                         (m2.from_user_id = m.to_user_id AND m2.to_user_id = m.from_user_id) 
                  )
ORDER BY m.created DESC

Answer 3

我使用GREATEST和LEAST来为每个对话创建一个grp。 然后对该grp进行排序并根据时间分配行号。

SQL DEMO

SELECT *
FROM (
        SELECT LEAST(`from_user_id`, `to_user_id`) as L,
               GREATEST(`from_user_id`, `to_user_id`) as G,
               `text`,
               CONCAT (LEAST(`from_user_id`, `to_user_id`), '-', GREATEST(`from_user_id`, `to_user_id`)) as grp,
               @rn := if(@grp = CONCAT(LEAST(`from_user_id`, `to_user_id`), '-', GREATEST(`from_user_id`, `to_user_id`)),
                         @rn + 1,
                         if(@grp := CONCAT(LEAST(`from_user_id`, `to_user_id`), '-', GREATEST(`from_user_id`, `to_user_id`)), 1, 1)
                         ) as rn,
               `time`
        FROM Table1
        CROSS JOIN (SELECT @rn := 0, @grp := '') as var
        ORDER BY LEAST(`from_user_id`, `to_user_id`),
                 GREATEST(`from_user_id`, `to_user_id`),
                 `time` DESC
     ) T
WHERE rn = 1;

OUTPUT

编辑：最后你需要过滤对话中的13。

WHERE rn = 1
  AND 13 IN (`L`, `G`);