select *,COUNT(feed_id) from
(SELECT feed_contents.*, feed.feed_url, feed.lang_direction, feed.feed_title
FROM feed_contents
INNER JOIN feed ON feed_contents.feed_id = feed.feed_id
INNER JOIN user_feeds ON feed_contents.feed_id = user_feeds.feed_id
WHERE user_feeds.user_id = 13
AND DATE(feed_contents.content_date) >= CURDATE() - INTERVAL 90 DAY
ORDER BY feed_contents.content_date desc) as tbl
group by feed_id
order by content_date desc
limit 0,20
我有这个查询来从多个表中获取结果， 结果它返回一个针对feed_id的记录， 我希望每个feed_is有5条记录
输出就像 http://screencast.com/t/HHxNOOSdSX4S
我希望每个
最多5个答案 0 :(得分：1)
您可以使用变量：
orderid变量"SELECT * FROM Orders ORDER BY orderid DESC LIMIT 20";
用于枚举每个SELECT feed_id, content_date, feed_url, lang_direction, feed_title
FROM (
SELECT feed_id, content_date, feed_url, lang_direction, feed_title,
@rn := IF(@fid = feed_id, @rn + 1,
IF(@fid := feed_id, 1, 1)) AS rn
FROM (
SELECT feed_contents.feed_id,
feed_contents.content_date,
feed.feed_url,
feed.lang_direction,
feed.feed_title
FROM feed_contents
INNER JOIN feed ON feed_contents.feed_id = feed.feed_id
INNER JOIN user_feeds ON feed_contents.feed_id = user_feeds.feed_id
WHERE user_feeds.user_id = 13 AND
DATE(feed_contents.content_date) >= CURDATE() - INTERVAL 90 DAY) AS tbl
CROSS JOIN (SELECT @rn := 0, @fid := 0) AS vars
ORDER BY feed_id, content_date DESC) AS s
WHERE s.rn <= 5
分区中的记录。 @rn值更改后，feed_id将重置为1，以便开始计算下一个分区。记录按字段feed_id按降序编号。