Question

我有两张桌子：

聊天（from只是发送了第一封邮件的人）

id | id_from | id_to
1  | 1       | 2
2  | 1       | 3
3  | 3       | 2
4  | 4       | 1

消息

id | content | date                | id_chat | id_from | id_to
1  | hi      | 2017-10-04 23:14:41 | 1       | 1       | 2
2  | hello   | 2017-10-04 23:15:03 | 1       | 2       | 1
3  | heey    | 2017-10-04 23:40:00 | 4       | 4       | 1

我想得到每次谈话的最后一条消息。

我试过（1是用户登录的ID）：

SELECT MAX(id) AS idMessage, id_from, id_to, content, date 
FROM message 
WHERE 1 IN (id_from, id_to) 
GROUP BY id_chat
ORDER BY idMessage 
DESC LIMIT 10

我收到了最新的对话，但我收到了每个对话的第一条消息，而不是最后一条消息。我怎样才能收到最后一条消息？

Answer 1

SELECT m.content as idMessage, m.id_from, m.id_to, m.date  from message as m JOIN chat c WHERE <Logged_ID> = m.id_chat ORDER BY m.date DESC LIMIT 1

Answer 2

select m1.*
from messages m1
join
(
  SELECT MAX(id) as id
  FROM message
  GROUP BY id_chat
) m2 on m1.id = m2.id

选择每个会话的最新消息

2 个答案: