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这是我的简单消息数据库

id | from_id | to_id | created_at

所以我有查询来获取对话框列表（这里我只从结果中获取 coll - 最后一条消息的最后一个 id 与 collocutor）：

SELECT IF(to_id = #{id}, from_id, to_id) as coll,
       LEAST(from_id,to_id), 
       GREATEST(from_id, to_id), 
       MAX(created_at), 
       MAX(id) FROM `messages`  
WHERE LEAST(from_id,to_id) = #{id} or GREATEST(from_id, to_id) = #{id}
GROUP BY LEAST(from_id,to_id), GREATEST(from_id, to_id)
ORDER BY MAX(created_at) DESC

从这个查询中，我得到了每个合作者的最后一条消息的列表。现在我需要获取所有最后一条消息的 id 与每个 colocutors 的最后一条消息是从 collocutor 到我的。第一个想法是通过 MAX(id) 加入消息表。于是代码诞生了：

SELECT IF( m1.to_id = #{id}, m1.from_id, m1.to_id ) AS coll,
       LEAST( m1.from_id, m1.to_id ) ,       
       GREATEST( m1.from_id, m1.to_id ) , 
       MAX( m1.created_at ) , 
       MAX( m1.id )
FROM  `messages` AS m1 
INNER  JOIN messages AS m2 on m2.id = MAX( m1.id )
WHERE  (
         LEAST( m1.from_id, m1.to_id ) = #{id}
         OR GREATEST( m1.from_id, m1.to_id ) = #{id}
       )
       AND
       m2.from <> #{id}
GROUP BY LEAST( m1.from_id, m1.to_id ) , GREATEST( m1.from_id, m1.to_id ) 
ORDER BY MAX( m1.created_at ) DESC

但是是上升#1111 错误。第二个想法是“嗯，实际上我需要将第二个消息表加入到第一个查询的结果中”。所以现在诞生了猴子代码：

Select * from (SELECT IF(to_id = #{id}, from_id, to_id) as coll,
       LEAST(from_id,to_id), 
       GREATEST(from_id, to_id), 
       MAX(created_at), 
       MAX(id) as max_id
FROM `messages`  
WHERE LEAST(from_id,to_id) = #{id} or GREATEST(from_id, to_id) = #{id}
GROUP BY LEAST(from_id,to_id), GREATEST(from_id, to_id)
ORDER BY MAX(created_at) DESC) m1
LEFT JOIN messages as m2 on m1.max_id = m2.id
WHERE to_id =#{id}

所以，这行得通。但是优化呢？有什么想法吗？

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