MySQL的左连接的时间太长

Question

我有以下SQL查询：

SELECT 
    upd.*, 
    usr.username AS `username`, 
    usr.profile_picture AS `profile_picture` 
FROM 
    updates AS upd 
LEFT JOIN 
    subscribers AS sub ON upd.uid=sub.suid 
LEFT JOIN 
    users AS usr ON upd.uid=usr.uid 
WHERE 
    upd.deleted='0' && (upd.uid='118697835834' || sub.uid='118697835834') 
GROUP BY upd.id 
ORDER BY upd.date DESC 
LIMIT 0, 15 

在那里我得到的所有用户（118697835834）的更新，使用左从另一个表他的个人资料图片加入以及所有他的订阅用户更新，所以我可以在他的新闻源中显示他们吗？

然而，随着更新，得到了越来越多这样的查询需要更多的时间来加载...现在使用CodeIgniter的探查，我可以看到的是，查询需要1.3793 ...

现在，我已经创建大约18K的虚拟账户和订阅我和反之亦然，所以我可以测试执行时间...我得到的时间是悲剧考虑到我在本地主机...

我也有一些索引，我想在用户表中需要更多（用户名和用户名唯一），更新表（update_id为唯一，用户名为索引）

我想我做错了什么让这么糟糕的结果...

编辑： 运行EXPLAIN EXTENDED结果：

Array 
(
    [0] => stdClass Object 
     (
      [id] => 1 
      [select_type] => SIMPLE 
      [table] => upd 
      [type] => ALL 
      [possible_keys] => i2 
      [key] => 
      [key_len] => 
      [ref] => 
      [rows] => 22 
      [filtered] => 100.00 
      [Extra] => Using where; Using temporary; Using filesort 
     ) 

    [1] => stdClass Object 
     (
      [id] => 1 
      [select_type] => SIMPLE 
      [table] => sub 
      [type] => ALL 
      [possible_keys] => 
      [key] => 
      [key_len] => 
      [ref] => 
      [rows] => 18244 
      [filtered] => 100.00 
      [Extra] => Using where 
     ) 

    [2] => stdClass Object 
     (
      [id] => 1 
      [select_type] => SIMPLE 
      [table] => usr 
      [type] => eq_ref 
      [possible_keys] => uid 
      [key] => uid 
      [key_len] => 8 
      [ref] => site.upd.uid 
      [rows] => 1 
      [filtered] => 100.00 
      [Extra] => 
     ) 

) 

EDIT2：SHOW创建表的 用户表：

CREATE TABLE `users` (
`id` bigint(20) NOT NULL AUTO_INCREMENT, 
`uid` bigint(20) NOT NULL, 
`username` varchar(20) COLLATE utf8_unicode_ci NOT NULL, 
`email` text CHARACTER SET latin1 NOT NULL, 
`password` text CHARACTER SET latin1 NOT NULL, 
`profile_picture_full` text COLLATE utf8_unicode_ci NOT NULL, 
`profile_picture` text COLLATE utf8_unicode_ci NOT NULL, 
`date_registered` datetime NOT NULL, 
`activated` tinyint(1) NOT NULL, 
`closed` tinyint(1) NOT NULL, 
PRIMARY KEY (`id`), 
UNIQUE KEY `uid` (`uid`), 
UNIQUE KEY `username` (`username`) 
) ENGINE=MyISAM AUTO_INCREMENT=23521 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci 

认购表：

个

CREATE TABLE `subscribers` (
`id` bigint(20) NOT NULL AUTO_INCREMENT, 
`sid` bigint(20) NOT NULL, 
`uid` bigint(20) NOT NULL, 
`suid` bigint(20) NOT NULL, 
`date` datetime NOT NULL, 
PRIMARY KEY (`id`) 
) ENGINE=MyISAM AUTO_INCREMENT=18255 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci 

更新表：

CREATE TABLE `updates` (
`id` int(11) NOT NULL AUTO_INCREMENT, 
`update_id` bigint(19) NOT NULL, 
`uid` bigint(20) NOT NULL, 
`type` text COLLATE utf8_unicode_ci NOT NULL, 
`update` text COLLATE utf8_unicode_ci NOT NULL, 
`date` datetime NOT NULL, 
`total_likes` int(11) NOT NULL, 
`total_comments` int(11) NOT NULL, 
`total_favorites` int(11) NOT NULL, 
`category` bigint(20) NOT NULL, 
`deleted` tinyint(1) NOT NULL, 
`deleted_date` datetime NOT NULL, 
PRIMARY KEY (`id`), 
UNIQUE KEY `i1` (`update_id`), 
KEY `i2` (`uid`), 
KEY `deleted_index` (`deleted`) 
) ENGINE=MyISAM AUTO_INCREMENT=23 DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci 

Answer 1

尝试这一个（不GROUP BY）：

SELECT 
    upd.*, 
    usr.username AS `username`, 
    usr.profile_picture AS `profile_picture` 
FROM 
     updates AS upd 
    LEFT JOIN 
     users AS usr 
      ON upd.uid = usr.uid 
WHERE 
    upd.deleted='0' 
    AND 
    (upd.uid='118697835834' 
    OR EXISTS 
     (SELECT * 
     FROM subscribers AS sub 
     WHERE upd.uid = sub.suid 
      AND sub.uid = '118697835834' 
    ) 
    ) 
ORDER BY upd.date DESC 
LIMIT 0, 15 

---

至少这在加入所用的列应被索引：updates.uid，users.uid和subscribers.suid。

我还会在subscribers.uid上添加一个索引。

Answer 2

应该用不了多长时间运行;你有'删除'索引吗？什么是'GROUP BY ID'在做什么？它应该是UID吗？它可以出来，如果ID实际上只是一个自动增量，唯一的ID？ （这将是昂贵的以及无意义）

Answer 3

尝试：

SELECT 
    upd.*, 
    usr.username AS `username`, 
    usr.profile_picture AS `profile_picture` 
FROM 
    updates AS upd 
LEFT JOIN 
    subscribers AS sub ON upd.uid=sub.suid 
LEFT JOIN 
    users AS usr ON upd.uid=usr.uid 
WHERE 
    upd.deleted=0 and upd.uid in (118697835834,118697835834) 
GROUP BY upd.id 
ORDER BY upd.date DESC 
LIMIT 0, 15 

注意'已经从数字值和位运算符除去变更为常规的运营商。

Answer 4

我认为你最好将这个查询分隔成用户表上的select，然后将这些结果与订户表上的select进行联合。

Answer 5

不使用连接，试试这个：

select *, 
     (select username from users where uid = upd.uid) as username, 
     (select profile_picture from users where uid = upd.uid) as profile_picture, 
from updates as upd 
WHERE 
    upd.deleted='0' && upd.uid='118697835834' 

（未测试！）

也许你得检查是否存在的where子句与另一子选择用户。

另一种方法是在子选择上进行连接，而不是在整个表上进行连接。这也可能会增加你的表现。