表:PHP的MySQL集团通过获得最新的记录,而不是第一个记录
(`post_id`, `forum_id`, `topic_id`, `post_time`)
(79, 8, 4, '2012-11-19 06:58:08');
(80, 3, 3, '2012-11-19 06:58:42'),
(81, 9, 9, '2012-11-19 06:59:04'),
(82, 11, 6, '2012-11-19 16:05:39'),
(83, 9, 9, '2012-11-19 16:07:46'),
(84, 9, 11, '2012-11-19 16:09:33'),
查询:
SELECT post_id, forum_id, topic_id FROM posts
GROUP BY topic_id
ORDER BY post_time DESC
LIMIT 5
结果:
[0] => [post_id] => 84 [forum_id] => 9 [topic_id] => 11
[1] => [post_id] => 82 [forum_id] => 11 [topic_id] => 6
[2] => [post_id] => 81 [forum_id] => 9 [topic_id] => 9
[3] => [post_id] => 80 [forum_id] => 3 [topic_id] => 3
[4] => [post_id] => 79 [forum_id] => 8 [topic_id] => 4
问题:
如何重写查询,以便我t返回post_id - > 83而不是post_id - > 81?
它们都具有相同的论坛和主题ID,但POST_ID - > 81比POST_ID的旧日期 - > 83.
但似乎集团通过获得的“第一”的记录,而不是“最新'一个。
我试图改变查询
SELECT post_id, forum_id, topic_id, MAX(post_time)
但同时返回POST_ID 81和83
感谢您的解释和代码。两个查询都有效。 – shanebp
我用过你第一个查询,但它在我改变 ...之前没有工作... WHERE post_time = ... TO ... WHERE post_time IN ...因为我得到了更多的结果。 –
这是为什么这么麻烦?使用SQL它将只是'选择*从[表] GROUP BY [列]'选择最新的,而不是最古老的,似乎合乎逻辑的权利? –