[英]Selecting records from a table and then selecting a count of records from another table
我有两张看起来像这样的桌子;
Table: pictures
`picture_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`picture_title` varchar(255) NOT NULL,
`picture_description` text NOT NULL,
`picture_src` varchar(50) NOT NULL,
`picture_filetype` varchar(10) NOT NULL,
`picture_width` int(11) NOT NULL,
`picture_height` int(11) NOT NULL,
`user_id` int(10) unsigned NOT NULL,
`upload_date` datetime NOT NULL,
-
Table: picture_votes
`vote_id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`picture_id` int(10) unsigned NOT NULL,
`vote` tinyint(4) NOT NULL,
`user_id` int(10) unsigned NOT NULL,
`timestamp` datetime NOT NULL,
我想要做的是从pictures
表中选择每个字段,然后在picture_votes
中所有记录的计数,其中pictures.picture_id = picture_votes.picture_id
,例如;
picture_id => 1
picture_title => 'Pic title'
picture_description => 'Pic description'
picture_src => 'b8b3f2c3a85f1a46fbf2ee132d81f783'
picture_filetype => 'jpg'
picture_width => 612
picture_height => 612
user_id => 1
upload_date => '2013-10-12 12:00:00'
vote_count => 3 // Amount of records in `picture_votes` that has `picture_id` = 1
我想出了(其中$limit
是要选择的图片数量);
SELECT pictures.*, count(picture_votes.vote) as vote_count
FROM pictures, picture_votes
WHERE pictures.picture_id = picture_votes.picture_id
ORDER BY upload_date DESC
LIMIT $limit
这仅选择1张图片和picture_votes
中所有记录的计数。
你真的想要使用LEFT join,因为这将返回所有图片,而不仅仅是投票的图片。 你也应该做总和(pv.vote)和COUNT(),如果你的选票超过1(嘿,它可能发生!只想:高级账户== x2票;-)
SELECT p.*, SUM(pv.vote) votes FROM pictures p
LEFT JOIN picture_votes pv
ON pv.picture_id=p.picture_id
GROUP BY pv.picture_id
如果您想排序前10位投票数:
SELECT * FROM (
SELECT p.*, SUM(pv.vote) votes FROM pictures p
LEFT JOIN picture_votes pv
ON pv.picture_id=p.picture_id
GROUP BY pv.picture_id
) AS aggregate
ORDER BY votes DESC
LIMIT 10;
如果picture_id
在picture_votes
表上编入索引,那么以下内容甚至可能比连接更快:
SELECT *
, ( SELECT COUNT(*) FROM picture_votes WHERE picture_id = pictures.picture_id )
FROM pictures
这可能会完全跳过表的内容,并允许您简单地计算哈希表中的记录,这些记录应该更快。
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