起因：在innodb表上做count(*)统计实在是太慢了，因此想办法看能不能再快点。
现象：先来看几个测试案例，如下
一、 sbtest 表上的测试
show create table sbtest\G *************************** 1. row *************************** Table: sbtest Create Table: CREATE TABLE `sbtest` ( `aid` bigint(20) unsigned NOT NULL auto_increment, `id` int(10) unsigned NOT NULL default '0', `k` int(10) unsigned NOT NULL default '0', `c` char(120) NOT NULL default '', `pad` char(60) NOT NULL default '', PRIMARY KEY (`aid`), KEY `k` (`k`), KEY `id` (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=1000001 DEFAULT CHARSET=latin1 show index from sbtest; +--------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | +--------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | sbtest | 0 | PRIMARY | 1 | aid | A | 1000099 | NULL | NULL | | BTREE | | | sbtest | 1 | k | 1 | k | A | 18 | NULL | NULL | | BTREE | | | sbtest | 1 | id | 1 | id | A | 1000099 | NULL | NULL | | BTREE | | +--------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
填充了 1000万条 记录。
1、 直接 count(*)
explain SELECT COUNT(*) FROM sbtest; +----+-------------+--------+-------+---------------+---------+---------+------+---------+-------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+--------+-------+---------------+---------+---------+------+---------+-------------+ | 1 | SIMPLE | sbtest | index | NULL | PRIMARY | 8 | NULL | 1000099 | Using index | +----+-------------+--------+-------+---------------+---------+---------+------+---------+-------------+ SELECT COUNT(*) FROM sbtest; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (1.42 sec)
可以看到，如果不加任何条件，那么优化器优先采用 primary key 来进行扫描。
2、count(*) 使用 primary key 字段做条件
explain SELECT COUNT(*) FROM sbtest WHERE aid>=0; +----+-------------+--------+-------+---------------+---------+---------+------+--------+--------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+--------+-------+---------------+---------+---------+------+--------+--------------------------+ | 1 | SIMPLE | sbtest | range | PRIMARY | PRIMARY | 8 | NULL | 485600 | Using where; Using index | +----+-------------+--------+-------+---------------+---------+---------+------+--------+--------------------------+ SELECT COUNT(*) FROM sbtest WHERE aid>=0; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (1.39 sec)
可以看到，尽管优化器认为只需要扫描 485600 条记录(其实是索引)，比刚才少多了，但其实仍然要做全表(索引)扫描。因此耗时和第一种相当。
3、 count(*) 使用 secondary index 字段做条件
explain SELECT COUNT(*) FROM sbtest WHERE id>=0; +----+-------------+--------+-------+---------------+------+---------+------+--------+--------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+--------+-------+---------------+------+---------+------+--------+--------------------------+ | 1 | SIMPLE | sbtest | range | id | id | 4 | NULL | 500049 | Using where; Using index | +----+-------------+--------+-------+---------------+------+---------+------+--------+--------------------------+ SELECT COUNT(*) FROM sbtest WHERE id>=0; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (0.43 sec)
可以看到，采用这种方式查询会非常快。
有人也许会问了，会不会是因为 id 字段的长度比 aid 字段的长度来的小，导致它扫描起来比较快呢？先不着急下结论，咱们来看看下面的测试例子。
二、 sbtest1 表上的测试
show create table sbtest1\G *************************** 1. row *************************** Table: sbtest1 Create Table: CREATE TABLE `sbtest1` ( `aid` int(10) unsigned NOT NULL AUTO_INCREMENT, `id` bigint(20) unsigned NOT NULL DEFAULT '0', `k` int(10) unsigned NOT NULL DEFAULT '0', `c` char(120) NOT NULL DEFAULT '', `pad` char(60) NOT NULL DEFAULT '', PRIMARY KEY (`aid`), KEY `k` (`k`), KEY `id` (`id`) ) ENGINE=InnoDB AUTO_INCREMENT=1000001 DEFAULT CHARSET=latin1 show index from sbtest1; +---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | Table | Non_unique | Key_name | Seq_in_index | Column_name | Collation | Cardinality | Sub_part | Packed | Null | Index_type | Comment | +---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+ | sbtest1 | 0 | PRIMARY | 1 | aid | A | 1000099 | NULL | NULL | | BTREE | | | sbtest1 | 1 | k | 1 | k | A | 18 | NULL | NULL | | BTREE | | | sbtest1 | 1 | id | 1 | id | A | 1000099 | NULL | NULL | | BTREE | | +---------+------------+----------+--------------+-------------+-----------+-------------+----------+--------+------+------------+---------+
这个表里，把 aid 和 id 的字段长度调换了一下，也填充了 1000万条 记录。
1、 直接 count(*)
explain SELECT COUNT(*) FROM sbtest1; +----+-------------+---------+-------+---------------+---------+---------+------+---------+-------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+---------+-------+---------------+---------+---------+------+---------+-------------+ | 1 | SIMPLE | sbtest1 | index | NULL | PRIMARY | 4 | NULL | 1000099 | Using index | +----+-------------+---------+-------+---------------+---------+---------+------+---------+-------------+ SELECT COUNT(*) FROM sbtest1; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (1.42 sec)
可以看到，如果不加任何条件，那么优化器优先采用 primary key 来进行扫描。
2、count(*) 使用 primary key 字段做条件
explain SELECT COUNT(*) FROM sbtest1 WHERE aid>=0; +----+-------------+---------+-------+---------------+---------+---------+------+--------+--------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+---------+-------+---------------+---------+---------+------+--------+--------------------------+ | 1 | SIMPLE | sbtest1 | range | PRIMARY | PRIMARY | 4 | NULL | 316200 | Using where; Using index | +----+-------------+---------+-------+---------------+---------+---------+------+--------+--------------------------+ 1 row in set (0.00 sec) SELECT COUNT(*) FROM sbtest1 WHERE aid>=0; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (1.42 sec)
可以看到，尽管优化器认为只需要扫描 485600 条记录(其实是索引)，比刚才少多了，但其实仍然要做全表(索引)扫描。因此耗时和第一种相当。
3、 count(*) 使用 secondary index 字段做条件
explain SELECT COUNT(*) FROM sbtest1 WHERE id>=0; +----+-------------+---------+-------+---------------+------+---------+------+--------+--------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+---------+-------+---------------+------+---------+------+--------+--------------------------+ | 1 | SIMPLE | sbtest1 | range | id | id | 8 | NULL | 500049 | Using where; Using index | +----+-------------+---------+-------+---------------+------+---------+------+--------+--------------------------+ 1 row in set (0.00 sec) SELECT COUNT(*) FROM sbtest1 WHERE id>=0; +----------+ | COUNT(*) | +----------+ | 1000000 | +----------+ 1 row in set (0.45 sec)
可以看到，采用这种方式查询会非常快。
上面的所有测试，均在 mysql 5.1.24 环境下通过，并且每次查询前都重启了 mysqld。
可以看到，把 aid 和 id 的长度调换之后，采用 secondary index 查询仍然是要比用 primary key 查询来的快很多。看来主要不是字段长度引起的索引扫描快慢，而是采用 primary key 以及 secondary index 引起的区别。那么，为什么用 secondary index 扫描反而比 primary key 扫描来的要快呢？我们就需要了解innodb的 clustered index 和secondary index 之间的区别了。
innodb 的 clustered index 是把 primary key 以及 row data 保存在一起的，而 secondary index 则是单独存放，然后有个指针指向 primary key。因此，需要进行 count(*) 统计表记录总数时，利用 secondary index 扫描起来，显然更快。而primary key则主要在扫描索引，同时要返回结果记录时的作用较大，例如：
SELECT * FROM sbtest WHERE aid = xxx;
那既然是使用 secondary index 会比 primary key 更快，为何优化器却优先选择 primary key 来扫描呢，Heikki Tuuri 的回答是：
in the example table, the secondary index is inserted into in a perfect order! That is very unusual. Normally the secondary index would be fragmented, causing random disk I/O, and the scan would be slower than in the primary index. I am changing this to a feature request: keep 'clustering ratio' statistics on a secondary index and do the scan there if the order is almost the same as in the primary index. I doubt this feature will ever be implemented, though.
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