php - 双左连接查询需要优化

Question

我一直在为 mysql 连接而苦苦挣扎，但已经开始整合更多内容，但尽管阅读了数十篇教程和 mysql 手册，但仍难以理解。

我的情况是我有3张 table :

/* 基本上是一个保存粉丝记录的表 */

    CREATE TABLE `fans` (
      `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
      `first_name` varchar(255) DEFAULT NULL,
      `middle_name` varchar(255) DEFAULT NULL,
      `last_name` varchar(255) DEFAULT NULL,
      `email` varchar(255) DEFAULT NULL,
      `join_date` datetime DEFAULT NULL,
      `twitter` varchar(255) DEFAULT NULL,
      `twitterCrawled` datetime DEFAULT NULL,
      `twitterImage` varchar(255) DEFAULT NULL,
      PRIMARY KEY (`id`),
      UNIQUE KEY `email` (`email`)
    ) ENGINE=MyISAM AUTO_INCREMENT=20413 DEFAULT CHARSET=latin1;

    /* A TABLE OF OUR TWITTER FOLLOWERS */

    CREATE TABLE `twitterFollowers` (
      `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
      `screenName` varchar(25) DEFAULT NULL,
      `twitterId` varchar(25) DEFAULT NULL,
      `customerId` int(11) DEFAULT NULL,
      `uniqueStr` varchar(50) DEFAULT NULL,
      PRIMARY KEY (`id`),
      UNIQUE KEY `unique` (`uniqueStr`)
    ) ENGINE=InnoDB AUTO_INCREMENT=13426 DEFAULT CHARSET=utf8;

    /* TABLE THAT SUGGESTS A LIKELY MATCH OF A TWITTER FOLLOWER BASED ON THE EMAIL / SCREEN NAME COMPARISON OF THE FAN vs OUR FOLLOWERS 
    IF SOMEONE (ie. a moderator) CONFIRMS OR DENIES THAT IT'S A GOOD MATCH THEY PUT A DATESTAMP IN `dismissed` */

    CREATE TABLE `contentSuggestion` (
      `id` int(11) unsigned NOT NULL AUTO_INCREMENT,
      `userId` int(11) DEFAULT NULL,
      `fanId` int(11) DEFAULT NULL,
      `twitterAccountId` int(11) DEFAULT NULL,
      `contentType` varchar(50) DEFAULT NULL,
      `contentString` varchar(255) DEFAULT NULL,
      `added` datetime DEFAULT NULL,
      `dismissed` datetime DEFAULT NULL,
      `uniqueStr` varchar(255) DEFAULT NULL,
      PRIMARY KEY (`id`),
      UNIQUE KEY `unstr` (`uniqueStr`)
    ) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=utf8;

What I'm trying to get is:

SELECT [fan columns] WHERE fan screen name IS IN twitterfollowers AND WHERE fan screen name IS NOT IN contentSuggestion (with a datestamp in dismissed)

My attempts so far:

~33 seconds

SELECT fans.id, tf.screenName as col1, tf.twitterId as col2 FROM fans LEFT JOIN twitterFollowers tf ON tf.screenName = fans.emailUsername LEFT JOIN contentSuggestion cs ON cs.contentString = tf.screenName WHERE dismissed IS NULL GROUP BY(fans.id) HAVING col1 != ''

~14 seconds

SELECT id, emailUsername FROM fans WHERE emailUsername IN(SELECT DISTINCT(screenName) FROM twitterFollowers) AND emailUsername NOT IN(SELECT DISTINCT(contentString) FROM contentSuggestion WHERE dismissed IS NULL) GROUP BY (fans.id);

9.53 seconds

SELECT fans.id, tf.screenName as col1, tf.twitterId as col2 FROM fans LEFT JOIN twitterFollowers tf ON tf.screenName = fans.emailUsername WHERE tf.uniqueStr NOT IN(SELECT uniqueStr FROM contentSuggestion WHERE dismissed IS NULL)

我希望有更好的方法。我一直在努力在单个 LEFT JOIN 之外真正使用 JOINS，这已经帮助我大大加快了其他查询的速度。

感谢你给与我的帮助。

Answer 1

我会选择第二种方法的变体。而不是 IN , 使用 EXISTS .然后添加正确的索引并删除聚合:

SELECT f.id, f.emailUsername
FROM fans f
WHERE EXISTS (SELECT 1
              FROM twitterFollowers tf
              WHERE f.emailUsername = tf.screenName
             ) AND
      NOT EXISTS (SELECT 1
                  FROM contentSuggestion cs
                  WHERE f.emailUsername = cs.contentString AND
                        cs.dismissed IS NULL
                 ) ;

然后确保您有以下索引:twitterFollowers(screenName)和 contentSuggestion(contentString, dismissed) .

一些注意事项:

使用 IN 时, 不要使用 SELECT DISTINCT .我不能 100% 确定 MySQL 总是足够聪明，可以忽略 DISTINCT。在子查询中(它是多余的)。

从历史上看，EXISTS比 IN 快在 MySQL 中。优化器在最近的版本中得到了改进。

为了提高性能，您需要正确的索引。
然后确保您有以下索引:twitterFollowers(screenName)和 contentSuggestion(contentString, dismissed) .

假设 fan.id是唯一的(一个非常合理的假设)，您不需要最终的 group by .