MySQL：与直接使用视图的底层JOIN进行查询相比，为什么使用VIEW的查询效率较低？

Question

我有三个表，bug，bugrule和bugtrace，它们之间的关系是：

bug     1--------N  bugrule
        id = bugid

bugrule 0---------N bugtrace
        id = ruleid

因为我几乎一直对bug <---> bugtrace之间的关系感兴趣，所以我创建了一个适当的VIEW，将其用作几个查询的一部分。有趣的是，使用此VIEW的查询的性能明显比使用基础JOIN的等效查询差。]

VIEW定义：

CREATE VIEW bugtracev AS
  SELECT t.*, r.bugid
      FROM bugtrace AS t
      LEFT JOIN bugrule AS r ON t.ruleid=r.id
    WHERE r.version IS NULL
使用VIEW的查询的执行计划（性能不佳：]

mysql> explain 
      SELECT c.id,state,
             (SELECT COUNT(DISTINCT(t.id)) FROM bugtracev AS t 
               WHERE t.bugid=c.id) 
       FROM bug AS c 
      WHERE c.version IS NULL
        AND c.id<10;
+----+--------------------+-------+-------+---------------+--------+---------+-----------------+---------+-----------------------+
| id | select_type        | table | type  | possible_keys | key    | key_len | ref             | rows    | Extra                 |
+----+--------------------+-------+-------+---------------+--------+---------+-----------------+---------+-----------------------+
|  1 | PRIMARY            | c     | range | id_2,id       | id_2   | 8       | NULL            |       3 | Using index condition |
|  2 | DEPENDENT SUBQUERY | t     | index | NULL          | ruleid | 9       | NULL            | 1426004 | Using index           |
|  2 | DEPENDENT SUBQUERY | r     | ref   | id_2,id       | id_2   | 8       | bugapp.t.ruleid |       1 | Using where           |
+----+--------------------+-------+-------+---------------+--------+---------+-----------------+---------+-----------------------+
3 rows in set (0.00 sec)
直接使用基础JOIN执行查询的计划（性能良好）：

mysql> explain 
       SELECT c.id,state,
              (SELECT COUNT(DISTINCT(t.id)) 
                 FROM bugtrace AS t
                 LEFT JOIN bugrule AS r ON t.ruleid=r.id 
                WHERE r.version IS NULL
                  AND r.bugid=c.id) 
        FROM bug AS c 
       WHERE c.version IS NULL
         AND c.id<10;
+----+--------------------+-------+-------+---------------+--------+---------+-------------+--------+-----------------------+
| id | select_type        | table | type  | possible_keys | key    | key_len | ref         | rows   | Extra                 |
+----+--------------------+-------+-------+---------------+--------+---------+-------------+--------+-----------------------+
|  1 | PRIMARY            | c     | range | id_2,id       | id_2   | 8       | NULL        |      3 | Using index condition |
|  2 | DEPENDENT SUBQUERY | r     | ref   | id_2,id,bugid | bugid  | 8       | bugapp.c.id |      1 | Using where           |
|  2 | DEPENDENT SUBQUERY | t     | ref   | ruleid        | ruleid | 9       | bugapp.r.id | 713002 | Using index           |
+----+--------------------+-------+-------+---------------+--------+---------+-------------+--------+-----------------------+
3 rows in set (0.00 sec)
[CREATE TABLE语句（由不相关的列减少）是：

mysql> show create table bug;
CREATE TABLE `bug` (
  `id` bigint(20) NOT NULL,
  `version` int(11) DEFAULT NULL,
  `state` varchar(16) DEFAULT NULL,
  UNIQUE KEY `id_2` (`id`,`version`),
  KEY `id` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

mysql> show create table bugrule;
CREATE TABLE `bugrule` (
  `id` bigint(20) NOT NULL,
  `version` int(11) DEFAULT NULL,
  `bugid` bigint(20) NOT NULL,
  UNIQUE KEY `id_2` (`id`,`version`),
  KEY `id` (`id`),
  KEY `bugid` (`bugid`),
  CONSTRAINT `bugrule_ibfk_1` FOREIGN KEY (`bugid`) REFERENCES `bug` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

mysql> show create table bugtrace;
CREATE TABLE `bugtrace` (
  `id` bigint(20) NOT NULL,
  `ruleid` bigint(20) DEFAULT NULL,
  PRIMARY KEY (`id`),
  KEY `ruleid` (`ruleid`),
  CONSTRAINT `bugtrace_ibfk_1` FOREIGN KEY (`ruleid`) REFERENCES `bugrule` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

我有三个表，bug，bugrule和bugtrace，它们之间的关系是：bug 1 -------- N bugrule id = bugid bugrule 0 --------- N bugtrace id = ruleid因为我快要走了...

Answer 1

您询问为什么

Answer 2

[使用MySQL 5.6（或更早版本）时，至少尝试使用MySQL 5.7。根据What’s New in MySQL 5.7?：