Question

在我的基于线程的消息传递系统中，表模式是

> messages table
id(int auto incr primary key)
body(varchar)
time(datetime)

>message_reference table
id(int auto incr primary key)
message_id(forgain key from message table)
sender
receiver

在这里，我想选择发送给新接收者的第一个消息ID，发件人是登录的用户。

使用多个查询和一些代码执行此操作显然是可行的，但可以通过单个查询来解决性能问题吗？

Answer 1

你可以尝试

编辑：

如果id是自动递增，那么id也会随着时间的推移而增加，您可以使用：

SELECT message_reference.message_id, message_reference.receiver, messages.body
FROM message_reference, messages
WHERE message_reference.message_id IN (SELECT  MIN(message_reference.message_id)
                            FROM message_reference
                            GROUP BY message_reference.receiver)
AND message_reference.message_id = messages.id AND message_reference.sender = <sender>

Answer 2

这是我对你想要的最好的猜测，但是如果你给出已知输入，示例数据和预期输出会更容易。

SELECT
    MR2.message_id
FROM (
    SELECT
        MR.sender,
        MR.receiver,
        M.MIN(`time`) AS min_time
    FROM
        Message_References MR -- Either use plural names (my personal preference) or singular, but don't mix them
    INNER JOIN Messages M ON
        M.id = MR.message_id
    WHERE
        MR.sender = <sender>
    GROUP BY
        MR.received) SQ
INNER JOIN Message_References MR2 ON
    MR2.sender = SQ.sender AND
    MR2.receiver = SQ.receiver AND
    MR2.`time` = SQ.min_time

Answer 3

select mr.message_id from
  message_reference as mr inner join 
    (select mr1.reciever max(m1.time) as time from messages as m1
       inner join message_reference as mr1 on mr1.message_id = m1.id
       group by mr1.reciever) as last
    on mr.reciever = last.reciever and mr.time = last.time

在接收者和时间上加入带有“每个接收者的最大时间”表的消息参考

Answer 4

嗯，我得到了答案，只是一组查询以我想要的方式工作。我用了查询

SELECT SENDER,
RECEIVER,
BODY,
TIME,
MESSAGE_ID
FROM MESSAGE_REF JOIN MESSAGE 
ON  MESSAGE.ID=MESSAGE_REF.MESSAGE_ID
ORDER BY 'TIME' GROUP BY RECEIVER`

感谢大家的帮助。

从表中选择具有不同列值的行

4 个答案: