Question

实际上我正在尝试创建一个像FB（消息）这样的会话界面，并且为此，使用sql查询来获取已经与用户交谈过的所有人。我需要他按照降序排列的用户的id，就像A与B和C聊天一样。然后B和C将是该查询的结果，B将首先出现因为A最近与B聊天。

我的'消息'表结构是： http://www.softnuke.com/me/files/DB.png

这是FB的例子： http://www.softnuke.com/me/files/msg.png

这是我的错误查询，需要修复：

SELECT DISTINCT(`mates`)FROM(
    SELECT `time` AS `time`,`from_id` AS `mates` 
     FROM `messages` AS T WHERE (`from_id`=$uid OR `to_id`=$uid) 

    UNION

    SELECT `time` AS `time`,`to_id` AS `mates`
    FROM `messages` AS T WHERE (`from_id`=$uid OR `to_id`=$uid) 

    ) AS T 
    WHERE `mates`!='$uid'
    ORDER BY `time`

$ uid会给我一个我想要获取List的用户的变量（这里是它的A）。

Answer 1

无论哪个人是主要人物，您似乎都在接触主要人物和与他们交谈的人。同样不太确定当你使用DISTINCT时，MySQL会如何计算订购时间，这会删除一些记录的时间。

您可以通过以下方式获得最长时间和顺序： -

SELECT `mates`, MAX(`time`) AS LatestConv
FROM(
    SELECT `time` AS `time`,`from_id` AS `mates` 
     FROM `messages` AS T WHERE `to_id`=$uid

    UNION

    SELECT `time` AS `time`,`to_id` AS `mates`
    FROM `messages` AS T WHERE `from_id`=$uid 

    ) AS T 
GROUP BY `mates`
ORDER BY LatestConv

获取最新消息的状态： -

SELECT a.mates, a.LatestConv, IFNULL(b.Status, c.Status)
FROM
(
    SELECT mates, MAX(`time`) AS LatestConv
    FROM(
        SELECT `time` AS `time`, from_id AS mates 
        FROM messages AS T 
        WHERE to_id = $uid
        UNION
        SELECT `time` AS `time`, to_id AS mates
        FROM messages AS T 
        WHERE from_id = $uid 
        ) AS T 
    GROUP BY `mates`
) a
LEFT OUTER JOIN messages b
ON a.mates = b.from_id AND a.LatestConv = b.`time` AND b.to_id = $uid
LEFT OUTER JOIN messages c
ON a.mates = c.to_id AND a.LatestConv = c.`time` AND c.from_id = $uid
ORDER BY LatestConv

请注意，如果同一个人的多条消息共享相同的最新时间，则可能会让人感到困惑。如果可能的话，可以按如下方式处理： -

SELECT a.mates, a.LatestConv, MAX(IFNULL(b.Status, c.Status))
FROM
(
    SELECT mates, MAX(`time`) AS LatestConv
    FROM(
        SELECT `time` AS `time`, from_id AS mates 
        FROM messages AS T 
        WHERE to_id = $uid
        UNION
        SELECT `time` AS `time`, to_id AS mates
        FROM messages AS T 
        WHERE from_id = $uid 
        ) AS T 
    GROUP BY `mates`
) a
LEFT OUTER JOIN messages b
ON a.mates = b.from_id AND a.LatestConv = b.`time` AND b.to_id = $uid
LEFT OUTER JOIN messages c
ON a.mates = c.to_id AND a.LatestConv = c.`time` AND c.from_id = $uid
GROUP BY a.mates, a.LatestConv
ORDER BY LatestConv

使用UNION和SORT的MySql查询

这是我的错误查询，需要修复：

1 个答案: