我做了一个声明，将返回一个字符串的数据显示用户收件箱只需输入他们的 user_id .
我有三个表：用户、对话和消息。我正在尝试生成一个sql语句，它将返回 conv_id , user_id ，name（来自users表），以及在该对话中发送的最后一条消息。到目前为止，我得出的结论是：

SELECT DISTINCT n.conversation AS conv_id, n.user_id, n.name, n.message_body FROM
    (SELECT CONCAT(u.first, " ", u.last) AS name, c.conv_id AS conversation, c.user_id, m.message_body
    FROM users AS u,
      (SELECT conv_id, CASE
        WHEN user1="14" THEN user2
        WHEN user2="14" THEN user1
        END AS user_id FROM conversations) AS c,
      (SELECT DISTINCT x.id, x.message_body, x.conv_id FROM messages AS x,
        (SELECT conv_id AS convs FROM conversations WHERE user1="14" or user2="14") AS y
      WHERE (x.message_from="14" or x.message_to="14") AND x.conv_id=y.convs
      ORDER by x.id DESC) AS m
    WHERE u.user_id=c.user_id) as n
  GROUP BY n.conversation

现在它返回 conversation , user_id ，和 name 我在寻找，但它给了我错误的 message_body 每次谈话。
我的table是这样的：
用户

user_id | first | last
14        John    Smith
15        Steve   Lowe
16        Will    Willis

对话

conv_id | user1 | user2
4         14     15
5         15     16
6         16     14

信息

id | conv_id | message_to | message_from | message_body
1    4         15           14             alpha
2    4         14           15             beta
3    4         15           14             charlie
4    4         14           15             delta
5    5         16           15             epsilon
6    5         15           16             foxtrot
7    5         16           15             golf
8    5         15           16             hotel
9    6         14           16             india
10   6         16           14             juliette
11   6         14           16             kilo
12   6         16           14             lima

为了 user_id =14我希望我的语句返回：

conv_id | user_id |     name     | message_body
4        15        Steve Lowe     delta
6        16        Will Willis    lima

这是一个很难理解的问题。也许我需要尝试使用连接。
请帮帮我！
sqlfiddle:http://sqlfiddle.com/#!9/70359d号