在此查询上使用RIGHT JOIN时JOIN非常慢

Question

我遇到了这个查询的问题,需要几秒钟才能完成.我已经尝试了很多优化,但此时我正在拍摄空白.

表格如下(并没有完全标准化,特别是轨道表)

CREATE TABLE `tracks` (
`id` int(14) unsigned NOT NULL AUTO_INCREMENT,
`artist` varchar(200) NOT NULL,
`track` varchar(200) NOT NULL,
`album` varchar(200) NOT NULL,
`path` text NOT NULL,
`tags` text NOT NULL,
`priority` int(10) NOT NULL DEFAULT '0',
`lastplayed` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
`lastrequested` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
`usable` int(1) NOT NULL DEFAULT '0',
`accepter` varchar(200) NOT NULL DEFAULT '',
`lasteditor` varchar(200) NOT NULL DEFAULT '',
`hash` varchar(40) DEFAULT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `hash` (`hash`),
FULLTEXT KEY `searchindex` (`tags`,`artist`,`track`,`album`),
FULLTEXT KEY `artist` (`artist`,`track`,`album`,`tags`)
) ENGINE=MyISAM AUTO_INCREMENT=3336 DEFAULT CHARSET=utf8

CREATE TABLE `esong` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT,
`hash` varchar(40) COLLATE utf8_bin NOT NULL,
`len` int(10) unsigned NOT NULL,
`meta` text COLLATE utf8_bin NOT NULL,
PRIMARY KEY (`id`),
UNIQUE KEY `hash` (`hash`)
) ENGINE=InnoDB AUTO_INCREMENT=16032 DEFAULT CHARSET=utf8 COLLATE=utf8_bin

CREATE TABLE `efave` (
`id` int(10) unsigned NOT NULL DEFAULT '0',
`inick` int(10) unsigned NOT NULL,
`isong` int(10) unsigned NOT NULL,
UNIQUE KEY `inick` (`inick`,`isong`),
KEY `isong` (`isong`),
CONSTRAINT `inick` FOREIGN KEY (`inick`) REFERENCES `enick` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
CONSTRAINT `isong` FOREIGN KEY (`isong`) REFERENCES `esong` (`id`) ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB DEFAULT CHARSET=utf8

CREATE TABLE `enick` (
`id` int(10) unsigned NOT NULL AUTO_INCREMENT
`nick` varchar(30) COLLATE utf8_bin NOT NULL,
`dta` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`dtb` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
PRIMARY KEY (`id`),
KEY `nick` (`nick`)
) ENGINE=InnoDB AUTO_INCREMENT=488 DEFAULT CHARSET=utf8 COLLATE=utf8_bin

我试图以正常速度执行的查询如下

SELECT esong.meta, tracks.id FROM tracks RIGHT JOIN esong ON tracks.hash = esong.hash JOIN efave ON efave.isong = esong.id JOIN enick ON efave.inick = enick.id WHERE enick.nick = lower('nickname');

如果你删除RIGHT JOIN并将其更改为JOIN,那么它很快

EXPLAIN给了我这个结果,似乎efave选择中存在一个小问题,但我不知道如何解决这个问题

+----+-------------+--------+--------+---------------+---------+---------+-----------------------+------+----------+--------------------------+
| id | select_type | table  | type   | possible_keys | key     | key_len | ref                   | rows | filtered | Extra                    |
+----+-------------+--------+--------+---------------+---------+---------+-----------------------+------+----------+--------------------------+
|  1 | SIMPLE      | enick  | ref    | PRIMARY,nick  | nick    | 92      | const                 |    1 |   100.00 | Using where; Using index |
|  1 | SIMPLE      | efave  | ref    | inick,isong   | inick   | 4       | radiosite.enick.id    |   12 |   100.00 | Using index              |
|  1 | SIMPLE      | esong  | eq_ref | PRIMARY       | PRIMARY | 4       | radiosite.efave.isong |    1 |   100.00 |                          |
|  1 | SIMPLE      | tracks | ALL    | hash          | NULL    | NULL    | NULL                  | 3210 |   100.00 |                          |
+----+-------------+--------+--------+---------------+---------+---------+-----------------------+------+----------+--------------------------+

Answer 1

你的解释看起来干净,脱颖而出,以我的唯一的事情是事实ESONG表使用的整理utf8_bin,和轨道表没有指定归类,这意味着它可能使用另一种排序规则类型.尝试对齐您的排序规则并查看连接的执行方式.