使用内部选择慢速 MySQL 连接

Question

以下查询有效，但随着 sendlog 表的大小随时间增加而变慢。目标是从 newsletter_subscribers 表中选择一个列表，其中没有在 newsletter_sendlog 表中具有给定时事通讯 ID 的电子邮件条目。目前，在我的 mysql 服务器上大约需要 2.2 秒，而 sendlog 中只有几千个条目。

SELECT `newsletter_subscribers`.* 
FROM `newsletter_subscribers`
    INNER JOIN `newsletter_to_subscriber` 
        ON newsletter_to_subscriber.subscriber_id = newsletter_subscribers.id
    LEFT JOIN (
        SELECT `newsletter_sendlog`.`subscriber_email` 
        FROM `newsletter_sendlog` 
        WHERE (newsletter_id='7')
      ) AS `sendlog` 
        ON newsletter_subscribers.email = sendlog.subscriber_email 
WHERE (sendlog.subscriber_email IS NULL) 
AND (newsletter_to_subscriber.newsletter_id = '7')

EXPLAIN(query) 输出以下内容：

我对 EXPLAIN 的输出不太熟悉，但如果我阅读正确，它会表明它没有使用我在 newsletter_sendlog.subscriber_email 上定义的索引。我试过在那个表上使用 USE INDEX(email) ，但它似乎没有生效。

关于如何优化它的任何建议？或者可能建议另一个相同的查询？

---

newsletter_sendlog 的创建表：

CREATE TABLE `newsletter_sendlog` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `subscriber_email` varchar(100) NOT NULL default '',
  `newsletter_id` int(11) default NULL,
  `sendstatus` int(11) default NULL,
  `senddate` timestamp NOT NULL default CURRENT_TIMESTAMP,
  PRIMARY KEY  (`id`),
  KEY `newsletter_id` (`newsletter_id`),
  KEY `email` (`subscriber_email`)
) ENGINE=MyISAM AUTO_INCREMENT=2933 DEFAULT CHARSET=latin1;

为 newsletter_subscribers 创建表：

CREATE TABLE `newsletter_subscribers` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `email` varchar(100) NOT NULL default '',
  `name` tinytext,
  PRIMARY KEY  (`id`),
  KEY `email` (`email`)
) ENGINE=MyISAM AUTO_INCREMENT=2964 DEFAULT CHARSET=utf8;

为 newsletter_to_subscriber 创建表：

CREATE TABLE `newsletter_to_subscriber` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `newsletter_id` int(11) NOT NULL,
  `subscriber_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`),
  KEY `newsletter_subscriber` (`newsletter_id`,`subscriber_id`)
) ENGINE=MyISAM AUTO_INCREMENT=2964 DEFAULT CHARSET=latin1;

更新：

在订阅者 ID 上添加索引后，为 newsletter_to_subscriber 创建表现在如下所示：

CREATE TABLE `newsletter_to_subscriber` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `newsletter_id` int(11) NOT NULL,
  `subscriber_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`),
  KEY `newsletter_subscriber` (`newsletter_id`,`subscriber_id`),
  KEY `subscriber` (`subscriber_id`)
) ENGINE=MyISAM AUTO_INCREMENT=2964 DEFAULT CHARSET=latin1;

@nobody 建议的查询的解释：

Answer 1

最好选择特定字段，而不是星号 (*)，并避免使用反引号 (`)。试试看下面的（重写的）查询是否效果更好：

SELECT 
    newsletter_subscribers.id,
    newsletter_subscribers.email,
    newsletter_subscribers.name
FROM
    newsletter_subscribers
    LEFT JOIN
        newsletter_to_subscriber
        ON
            newsletter_to_subscriber.subscriber_id = newsletter_subscribers.id
    LEFT JOIN
        newsletter_sendlog
        ON
            newsletter_subscribers.email = newsletter_sendlog.subscriber_email
WHERE
    newsletter_to_subscriber.newsletter_id = 7
    AND
        newsletter_sendlog.newsletter_id = 7
    AND
        newsletter_sendlog.subscriber_email IS NULL

Answer 2

SELECT `newsletter_subscribers`.* FROM `newsletter_subscribers`
  INNER JOIN `newsletter_to_subscriber` 
    ON newsletter_to_subscriber.subscriber_id = newsletter_subscribers.id
  LEFT JOIN (
      SELECT `newsletter_sendlog`.`subscriber_email` FROM `newsletter_sendlog` 
        WHERE (newsletter_id='7')) AS `sendlog` 
    ON newsletter_subscribers.email=sendlog.subscriber_email 
  WHERE (sendlog.subscriber_email IS NULL) 
    AND (newsletter_to_subscriber.newsletter_id = '7')

你可以尝试在单列上实现索引键newsletter_to_subscriber.subscriber_id

看看有没有用？

尝试使用如下表结构：

CREATE TABLE `newsletter_to_subscriber` (
  `id` int(11) unsigned NOT NULL auto_increment,
  `newsletter_id` int(11) NOT NULL,
  `subscriber_id` int(11) NOT NULL,
  PRIMARY KEY  (`id`),
  KEY `newsletter_subscriber` (`newsletter_id`,`subscriber_id`)
  KEY `subscriber_id_key` (`subscriber_id`)
  KEY `newsletter_id_key` (`newsletter_id`)
) ENGINE=MyISAM AUTO_INCREMENT=2964 DEFAULT CHARSET=latin1;

Answer 3

不完全确定，但我认为索引被忽略，因为您正在寻找 NULL 值。

虽然有一种不同的、希望更有效的方式来运行此查询：

select *
from newsletter_subscribers
where email not in 
(select subscriber_email
from newsletter_sendlog
where newsletter_id='7')

Answer 4

首先你不需要那个子查询：

SELECT `newsletter_subscribers`.* 
FROM `newsletter_subscribers`
    INNER JOIN `newsletter_to_subscriber` 
        ON( newsletter_to_subscriber.subscriber_id = newsletter_subscribers.id )
    LEFT JOIN `newsletter_sendlog`
        ON( newsletter_subscribers.email = newsletter_sendlog.subscriber_email AND
            newsletter_sendlog.newsletter_id = '7' )
WHERE newsletter_sendlog.subscriber_email IS NULL

上面的查询将完成这项工作。

其次，在newsletter_to_subscriber 中，您在newsletter_id 和subscriber_id 上有一个多部分索引，您的查询不能使用它，因为它将搜索subscriber_id，并且在索引中排在第二位，您需要在subscriber_id 上有一个单独的索引：

INDEX( subscriber_id )