blocks|key|566320|text|你可以试试|type|unstyled|depth|inlineStyleRanges|entityRanges|data|566321|编辑：|566322|如果id是自动递增的，那么id也会随时间增加，您可以使用：|566323|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>|code-block|syntax|javascript|566324|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|O|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|P|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can try

EDIT:

If the id is auto increment, then the id will also increase with time and you can use:

<pre><code>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 = &lt;sender&gt;
</code></pre>

blocks|key|4565959|text|下面是我对你想要什么的最好的猜测，但是如果你给出了已知的输入，示例数据和期望的输出，它会更容易。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4565960|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|code-block|syntax|javascript|4565961|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here's my best guess as to what you want, but it would be easier if you gave known inputs, example data, and expected output.

<pre><code>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 = &lt;sender&gt;
 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
</code></pre>

blocks|key|566374|text|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|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|566375|将消息引用与接收者和时间上的“每个接收者的最大时间”表连接起来|unstyled|566376|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>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
</code></pre>

join message reference with "maxtime per reciever" table on reciever and time

blocks|key|2666860|text|我得到了答案，只是group+by+query以我想要的方式工作。我使用了query|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2666861|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`|code-block|syntax|javascript|2666862|感谢大家的帮助。|2666863|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Well I got the answer, Just a group by query worked the way I wanted. I used query 

<pre><code>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`
</code></pre>

Thanks everyone for the help.

In my thread based messaging system, the table schema is 

<pre><code>&gt; messages table
id(int auto incr primary key)
body(varchar)
time(datetime)

&gt;message_reference table
id(int auto incr primary key)
message_id(forgain key from message table)
sender
receiver
</code></pre>

Here, I want to select the first message id which is sent to a new receiver and sender is the user who is logged in.

Doing this with multiple queries and some code is obviously possible but can it be done with a single query for performance issues??

selecting rows from table which have distinct values for a column

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在我的基于线程的消息传递系统中，表模式是> messages tableid(int auto incr primary key)body(varchar)time(datetime)>message_reference tableid(int auto incr primary key)message_id(forg...

问从表中选择某一列具有不同值的行
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