SQL：如何按两列的唯一组合进行分组？

Question

上下文：

A table message 有列 from_user_id 和 to_user_id
用户应该看到最近的对话以及显示的最后一条消息
一个会话由多条消息组成，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）

Table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
+-------------------------------------------------+--------------+------------+

我尝试按from_user_id和to_user_id进行分组，但我显然得到了一组用户收到的消息和另一组用户发送的消息。

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 |
+-------------------------------+--------------+------------+---------------------+

虽然我想要：

+-------------------------------+--------------+------------+---------------------+
| 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。它确实对条目进行了正确分组，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

做你想做的事情的一种方法是使用相关子查询，找到为匹配对话创建的最小值 date/time：

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 2

我使用 GREATEST 和 LEAST 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;

输出

编辑：最后你需要从对话中过滤掉 13。

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

Answer 3

与#13 的最后一次对话？在更新的 DBMS 中，您将使用 row_number() 来查找这些。在 MySQL 中你可以使用 not exists，以确保对话伙伴没有更晚的 post。顺便说一句，您可以使用 from_user_id + to_user_id - 13 轻松找到合作伙伴的号码。（而当比较两条记录时，你可以直接使用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
);

SQL：如何按两列的唯一组合进行分组？

SQL: How do I group by a unique combination of two columns?

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