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我有这3张桌子
forums_forum
+-----+--------+-------------+-------+-----+
| fid | name | description | index | cid |
+-----+--------+-------------+-------+-----+
| 36 | gdghdf | hjghj | 54 | 5 |
| 45 | yutuy | iuyi | 99 | 6 |
+-----+--------+-------------+-------+-----+
forums_threads
+----+-----+-----+-------+-------+------+-----------+------+
| id | tid | fid | moved | mfrom | view | important | lock |
+----+-----+-----+-------+-------+------+-----------+------+
| 1 | 4 | 36 | 0 | NULL | 0 | 0 | 0 |
| 2 | 12 | 36 | 0 | NULL | 7 | 0 | 0 |
| 3 | 9 | 15 | 0 | NULL | 0 | 0 | 0 |
+----+-----+-----+-------+-------+------+-----------+------+
forums_posts
+----+-------+--------+--------+---------------------+--------+--------+-----+
| id | title | detail | author | date | edited | editby | tid |
+----+-------+--------+--------+---------------------+--------+--------+-----+
| 1 | asfsd | sdfsd | 1 | 2010-07-01 21:31:29 | 0 | NULL | 4 |
+----+-------+--------+--------+---------------------+--------+--------+-----+
我正在尝试创建返回结果的查询 – >对于每个独特的’fid’,来自’forums_posts’的一行(ORDER BY’日期’).
forums_forum.fid = forums_threads.fid
forums_threads.tid = forums_posts.tid
谢谢
最佳答案
这是Stack Overflow上经常出现的令人尊敬的最大n组问题.这是给出表格的解决方案：
SELECT p.* FROM forums_posts p JOIN forums_threads t ON p.tid = t.tid
WHERE NOT EXISTS (
SELECT * FROM forums_posts p2 JOIN forums_threads t2 ON p2.tid = t2.tid
WHERE t.fid = t2.fid AND p.date < p2.date
);
（编辑：李大同）
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