Table ideas: id, author_id, some_columns
Table ideas_tags: idea_id, tag_name
Table ideas_seen: idea_id, user_id
Table user: uid, ban, some_columns我需要从列表中得到10个有标记的想法,它们的作者没有被禁止,并且不是在ideas_seen中为当前的用户。
现在,我的查询如下:
SELECT
ideas.*, GROUP_CONCAT(DISTINCT IT_V.tag_name SEPARATOR '|||') AS tags
FROM `ideas`
LEFT JOIN ideas_tags IT
ON ideas.id=IT.idea_id
LEFT JOIN ideas_tags IT_V
ON ideas.id=IT_V.idea_id
LEFT JOIN ideas_seen IV
ON ideas.id=IV.idea_id AND IV.user_id=145974517
LEFT JOIN users ON users.uid=ideas.author_id
WHERE author_id!=145974517 AND IV.id IS NULL AND ( (IT.tag_name = 'some_tag') OR (IT.tag_name = 'another_tag') OR (IT.tag_name IS NULL) ) AND active=1 AND deleted=0 AND (users.ban=0 OR users.ban IS NULL)
GROUP BY ideas.id
ORDER BY id DESC
LIMIT 10这是网站上最慢的查询,我不知道如何加快速度。
解释:

CREATE TABLE IF NOT EXISTS `ideas` (
`id` int(11) NOT NULL,
`tutorial` tinyint(4) NOT NULL,
`text` text NOT NULL,
`author_id` int(11) NOT NULL,
`active` bit(1) NOT NULL,
`timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`views` int(11) NOT NULL,
`views_all` int(11) DEFAULT '0',
`deleted` tinyint(4) NOT NULL DEFAULT '0',
`many_users` tinyint(4) DEFAULT NULL,
`game_id` int(11) DEFAULT NULL
) ENGINE=InnoDB AUTO_INCREMENT=35983 DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `ideas_seen` (
`id` int(11) NOT NULL,
`idea_id` int(11) NOT NULL,
`user_id` int(11) NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=3694368 DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `ideas_tags` (
`id` int(11) NOT NULL,
`idea_id` int(11) NOT NULL,
`tag_name` tinytext NOT NULL
) ENGINE=InnoDB AUTO_INCREMENT=86832 DEFAULT CHARSET=utf8;
CREATE TABLE IF NOT EXISTS `users` (
`uid` int(11) NOT NULL,
`email` tinytext,
`password_hash` tinytext,
`restore_code` tinytext NOT NULL,
`last_action` timestamp NULL DEFAULT NULL,
`score` float NOT NULL,
`date_register` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`posts_length` int(11) DEFAULT NULL,
`settings` text,
`titles` int(11) NOT NULL DEFAULT '1',
`filter` int(11) NOT NULL DEFAULT '1',
`note` text NOT NULL,
`ban` tinyint(4) DEFAULT NULL,
`mod_send` smallint(6) DEFAULT '0',
`mod_get` int(11) DEFAULT '0',
`fp_notified` int(11) NOT NULL DEFAULT '0',
`skilled` tinyint(4) NOT NULL DEFAULT '0',
`show_only_skilled` tinyint(4) NOT NULL DEFAULT '0'
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
ALTER TABLE `ideas`
ADD PRIMARY KEY (`id`), ADD KEY `author_id` (`author_id`), ADD KEY `game_id` (`game_id`), ADD KEY `active` (`active`), ADD KEY `many_users` (`many_users`), ADD KEY `deleted` (`deleted`), ADD KEY `tutorial` (`tutorial`), ADD FULLTEXT KEY `idea_text` (`text`);
ALTER TABLE `ideas_seen`
ADD PRIMARY KEY (`id`), ADD KEY `user_id` (`user_id`), ADD KEY `idea_id` (`idea_id`);
ALTER TABLE `ideas_tags`
ADD PRIMARY KEY (`id`), ADD KEY `tag_name` (`tag_name`(255)), ADD KEY `idea_id` (`idea_id`);
ALTER TABLE `users`
ADD PRIMARY KEY (`uid`), ADD UNIQUE KEY `email` (`email`(255)), ADD KEY `ban` (`ban`), ADD KEY `fp_notified` (`fp_notified`), ADD KEY `skilled` (`skilled`);
ALTER TABLE `ideas`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=35983;
ALTER TABLE `ideas_seen`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=3694368;
ALTER TABLE `ideas_tags`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT,AUTO_INCREMENT=86832;
ALTER TABLE `ideas`
ADD CONSTRAINT `ideas_ibfk_1` FOREIGN KEY (`game_id`) REFERENCES `games` (`id`) ON DELETE NO ACTION;
ALTER TABLE `ideas_seen`
ADD CONSTRAINT `ideas_seen_ibfk_1` FOREIGN KEY (`idea_id`) REFERENCES `ideas` (`id`) ON DELETE CASCADE ON UPDATE NO ACTION;
ALTER TABLE `ideas_tags`
ADD CONSTRAINT `ideas_tags_ibfk_2` FOREIGN KEY (`idea_id`) REFERENCES `ideas` (`id`) ON DELETE CASCADE ON UPDATE NO ACTION;发布于 2017-05-23 22:37:34
TINYTEXT;改为VARCHAR(255),去掉索引上的“前缀”。也就是说,将INDEX email(255)更改为INDEX(email)。deletedPRIMARY KEY(id)和INDEX(id, ...)。在PRIMARY KEY的情况下,保留它;将另一个抛出,因为它没有提供额外的好处。idea_tags,看看是否可以避免这种情况。JOINs来膨胀行数,然后使用GROUP BY返回到原来的行(较少一些被过滤掉的行)。在这样做的过程中,庞大的ideas.*会在临时表中来回移动。TEXT (包括TINYTEXT)阻止了对tmp表更有效地使用MEMORY。让我们走过去,消除膨胀-排气。
首先,让我们来构造外部部分:
SELECT ideas.*, ( ??? ) as tags
FROM ideas
WHERE ???
ORDER BY ideas.id DESC
LIMIT 10;假设我们可以填写???,那么我们现在有了一条更快的评估路径。这将需要在id上建立索引,这是您在PRIMARY KEY(id)中所拥有的。幸运的话(没有WHERE),只需要触摸10行。(在您的版本中,必须收集、分组、排序整个表,然后才交付10个表。)
由于您的所有JOINs都是LEFT JOINs,因此可以证明,与ideas以外的表有关的WHERE子句不会过滤掉任何行。只剩下
WHERE author_id!=145974517
AND active=1
AND deleted=0为此,让我们使用(虽然我不确定它是否会被使用):
INDEX(active, deleted, author_id)回到AS tags..。现在,去掉查询,只获取tag_name值,为给定的ideas.id生成GROUP_CONCAT
SELECT GROUP_CONCAT(DISTINCT IT_V.tag_name SEPARATOR '|||') AS tags
FROM ideas_tags IT_V ON ideas.id = IT_V.idea_id
AND ( IT.tag_name = 'some_tag'
OR IT.tag_name = 'another_tag'
OR IT.tag_name IS NULL
)(这就是为什么有两个连接到idea_tags的原因。)同时,我建议SELECT可以作为获取tags的子查询。
嗯..。关于
LEFT JOIN users ON users.uid = ideas.author_id
WHERE ( users.ban=0 OR users.ban IS NULL )这似乎没有过滤,因为它是LEFT。它似乎没有提供任何列,因为在其他地方没有提到users。那么,我必须假设它是浪费代码?
同为
LEFT JOIN ideas_seen IV ON ideas.id = IV.idea_id
AND IV.user_id=145974517
WHERE IV.id IS NULL因此,它归结为删除一些索引,添加一个索引,并将查询重写为
SELECT ideas.*,
( SELECT GROUP_CONCAT(DISTINCT IT_V.tag_name SEPARATOR '|||')
FROM ideas_tags IT_V
WHERE ideas.id = IT_V.idea_id
AND ( IT.tag_name = 'some_tag'
OR IT.tag_name = 'another_tag'
OR IT.tag_name IS NULL )
) as tags
FROM ideas
WHERE author_id!=145974517
AND active=1
AND deleted=0
ORDER BY ideas.id DESC
LIMIT 10;也许,DISTINCT是不必要的。
https://stackoverflow.com/questions/44134499
复制相似问题