4

这是我的两张表:

TABLE: friends

'id' // primary id, auto increment
'uid' // user id who sent friend request
'fid' // friend id who received friend request
'status' //status of friendship, 1=active

.

TABLE: messages

'id' // message id (primary id), auto increment
'uid' // user who sent the message
'fid' // friend who received the message
'time' // time when the message was sent
'status' // status of message, read or unread.

我只想显示我发送消息或接收消息的朋友列表,按最后一次发送消息的时间(由朋友或我)排序。一位朋友应该只列出一次。我该怎么做?

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1 回答 1

4

我只想显示我发送消息或接收消息的朋友列表,按最后一次发送消息的时间(由朋友或我)排序。

尝试这个:

SELECT DISTINCT friends.id
FROM messages m
INNER JOIN
(
    SELECT uid id FROM friends WHERE status = 1 AND fid = myuserid
    UNION ALL
    SELECT fid FROM friends WHERE status = 1 AND uid = myuserid
) friends ON m.fid = friends.id OR m.uid = friends.id

但是,如果有一个users表,你可以这样做:

SELECT 
  senders.name 'From', 
  recievers.name 'To', 
  m.id, 
  m.body, 
  m.messagetime,
  m.status
FROM messages m
INNER JOIN
(
    SELECT uid id FROM friends WHERE status = 1 AND fid = 1
    UNION ALL
    SELECT fid    FROM friends WHERE status = 1 AND uid = 1
) friends ON m.fid = friends.id OR m.uid = friends.id
INNER JOIN users senders ON m.uid = senders.id
INNER JOIN users recievers ON m.fid = recievers.id
WHERE m.uid = 1 
   OR m.fid = 1
ORDER BY m.messagetime ASC OR DESC

SQL 小提琴演示

例如,这将为您提供:

| FROM | TO | ID |   BODY | MESSAGETIME | STATUS |
--------------------------------------------------
|   Me |  B |  1 |  hiiii |  2012-12-01 |      1 |
|    c | Me |  7 | sadfds |  2012-12-01 |      1 |
|   Me |  B |  8 |    ddd |  2012-12-10 |      1 |

这个查询是如何工作的?

查询:

SELECT uid id FROM friends WHERE status = 1 AND fid = myuserid
UNION ALL
SELECT fid    FROM friends WHERE status = 1 AND uid = myuserid

会给你你朋友的名单,你的朋友是:

  • 用户向您发送了友谊请求并接受了,或者
  • 用户向他发送友谊请求,他接受了。

这就是为什么我在您的用户 ID 中使用 a 的原因UNION ALLfid =并且我还假设这status = 1意味着友谊请求被接受。

这些是你的朋友。然后,要获取您已向其发送消息或从其接收消息的朋友列表,我们必须将此结果集与messages表连接起来。但是要获取发送给您的消息或您发送的消息,我们必须选择加入条件m.fid = friends.id OR m.uid = friends.id。而已。

于 2012-12-09T12:53:22.783 回答