Question

SELECT s1.receiverID, s2.senderID
FROM messages as s1
LEFT JOIN messages as s2 ON s1.receiverID = s2.senderID

我想组合“messages”表，其中senderID和receiverID是相同的，以实现像会话流一样的东西，但是当我运行它时，它返回总共10,422行。当所有行都是3000时，它如何返回更多？！

//示例场景：id = 8的用户访问此url sample.com/inbox/user=275。我希望显示它们之间的所有消息，以实现对话流/聊天等功能。

Answer 1

你不想加入。你只想获得所有用户都是发送者和接收者的消息。

select * -- (or whatever)
from messages
where
(senderid = [userId] and receiverid = [otherUserId])
or (senderid = [otherUserId] and receiverid = [userId])

你写的连接是说“给我系统中的每个消息组合，其中一条消息的发送者与另一条消息的接收者相同，如果有人收到了消息但没有发送任何消息，给我看那些带有空值的发送者值。“

Answer 2

也许这就是你真正想要的东西？

select t1.senderID as t1senderID, t1.receiverID as t1receiverID,
  t2.senderID as t2senderID, t2.receiverID as t2receiverID
from messages t1
join messages t2
on t1.senderID = t2.receiverID and t1.receiverID = t2.senderID

样本日期会有很大帮助。

修改

鉴于此评论：

我们有用户A和用户B，我想要返回它们之间的所有消息

你正在寻找这个：

select * from t1 where (senderID = 'A' and receiverID = 'B') or (senderID = 'B' and receiverID = 'A')

Answer 3

此示例可帮助您找出在连接后获得更多行的原因。

rID sID
1   2
1   3
2   1
3   1

s1.rID s1.sID s2.rID s2.sID
1      2      2      1
1      2      3      1
1      3      2      1
1      3      3      1
2      1      1      2
3      1      1      3

左连接一个表自己

3 个答案: