从每个对话中获取最后一条消息

Question

我之前曾经问过类似的问题,但是没有一个问题具有相同的条件,他们的答案对于这种情况不起作用.

包含消息的表如下所示:

id | owner_id | recipient_id | content      | created
 1 |        1 |            2 | Hello        | 2015-12-08 20:00
 2 |        2 |            1 | Hey          | 2015-12-08 20:10
 3 |        3 |            1 | You there?   | 2015-12-08 21:00
 4 |        1 |            3 | Yes          | 2015-12-08 21:15
 5 |        4 |            1 | Hey buddy    | 2015-12-08 22:00

让我们说我查询用户ID 1的对话,预期结果是:

id | owner_id | recipient_id | content      | created
 5 |        4 |            1 | Hey buddy    | 2015-12-08 22:00
 4 |        1 |            3 | Yes          | 2015-12-08 21:15
 2 |        2 |            1 | Hey          | 2015-12-08 20:10

我尝试了许多组合,使用JOIN和子查询,但没有一个给出预期的结果.

我正在寻找MySQL的答案,但这将用于CakePHP 2下开发的项目,所以如果有一个特定于Cake的方法,那就太好了.

这是我尝试过的一个查询,但它不起作用.我相信甚至不能满足我的需求.

SELECT
    IF ( owner_id = 1, recipient_id, owner_id ) AS Recipient,

    (
        SELECT
            content
        FROM
            messages

        WHERE
            ( owner_id = 1 AND recipient_id = Recipient  )
        OR
            ( owner_id = Recipient AND recipient_id = 1 )

        ORDER BY
            created DESC

        LIMIT 1
    )
FROM
    messages

WHERE
    owner_id = 1
OR
    recipient_id = 1

GROUP BY
    Recipient;

Answer 1

select t.* 
    from 
        t 
      join 
        (select user, max(created) m  
            from 
               (
                 (select id, recipient_id user, created 
                   from t 
                   where owner_id=1 ) 
               union 
                 (select id, owner_id user, created
                   from t 
                   where recipient_id=1)
                ) t1
           group by user) t2
     on ((owner_id=1 and recipient_id=user) or 
         (owner_id=user and recipient_id=1)) and 
         (created = m)
   order by created desc

在sqlfiddle上的例子