1. 问题描述
群友提出问题,表里有两个列c1、c2,分别为INT、VARCHAR类型,且分别创建了unique key。
SQL查询的条件是 WHERE c1 = ? AND c2 = ?
,用EXPLAIN查看执行计划,发现优化器优先选择了VARCHAR类型的c2列索引。
他表示很不理解,难道不应该选择看起来代价更小的INT类型的c1列吗?
2. 问题复现
创建测试表t1:
[root@yejr.run]> CREATE TABLE `t1` (
`c1` int NOT NULL AUTO_INCREMENT,
`c2` int unsigned NOT NULL,
`c3` varchar(20) NOT NULL,
`c4` varchar(20) NOT NULL,
PRIMARY KEY (`c1`),
UNIQUE KEY `k3` (`c3`),
UNIQUE KEY `k2` (`c2`)
) ENGINE=InnoDB;
利用 mysql_random_data_load
写入一万行数据:
mysql_random_data_load -h127.0.0.1 -uX -pX yejr t1 10000
查看执行计划:
[root@yejr.run]> EXPLAIN SELECT * FROM t1 WHERE
c2 = 1755950419 AND c3 = 'MichaelaAnderson'\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: t1
partitions: NULL
type: const
possible_keys: k3,k2
key: k3
key_len: 82
ref: const
rows: 1
filtered: 100.00
Extra: NULL
可以看到优化器的确选择了 k3 索引,而非"预期"的 k2 索引,这是为什么呢?
3. 问题分析
其实原因很简单粗暴:优化器认为这两个索引选择的代价都是一样的,只是优先选中排在前面的那个索引而已。
再建一个相同的表 t2,只不过把 k2、k3 的索引创建顺序对调下:
[root@yejr.run]> CREATE TABLE `t2` (
`c1` int NOT NULL AUTO_INCREMENT,
`c2` int unsigned NOT NULL,
`c3` varchar(20) NOT NULL,
`c4` varchar(20) NOT NULL,
PRIMARY KEY (`c1`),
UNIQUE KEY `k2` (`c2`),
UNIQUE KEY `k3` (`c3`)
) ENGINE=InnoDB;
再查看执行计划:
[root@yejr.run]> EXPLAIN SELECT * FROM t2 WHERE
c2 = 1755950419 AND c3 = 'MichaelaAnderson'\G
*************************** 1. row ***************************
id: 1
select_type: SIMPLE
table: t1
partitions: NULL
type: const
possible_keys: k2,k3
key: k2
key_len: 4
ref: const
rows: 1
filtered: 100.00
Extra: NULL
我们利用 EXPLAIN ANALYZE
来查看下两次执行计划的代价对比:
-- 查看t1表执行计划代价
[root@yejr.run]> EXPLAIN ANALYZE SELECT * FROM t1 WHERE
c2 = 1755950419 AND c3 = 'MichaelaAnderson'\G
*************************** 1. row ***************************
EXPLAIN: -> Rows fetched before execution (cost=0.00..0.00 rows=1) (actual time=0.000..0.000 rows=1 loops=1)
-- 查看t2表执行计划代价
[root@yejr.run]> EXPLAIN ANALYZE SELECT * FROM t2 WHERE c2 = 1755950419 AND c3 = 'MichaelaAnderson'\G
*************************** 1. row ***************************
EXPLAIN: -> Rows fetched before execution (cost=0.00..0.00 rows=1) (actual time=0.000..0.000 rows=1 loops=1)
可以看到,很明显代价都是一样的。
再利用 OPTIMIZE_TRACE
查看执行计划,也能看到两个SQL的代价是一样的:
...
{
"rows_estimation": [
{
"table": "`t1`",
"rows": 1,
"cost": 1,
"table_type": "const",
"empty": false
}
]
},
...
所以,优化器认为选择哪个索引都是一样的,就看哪个索引排序更靠前。
从执行SELECT时的debug trace结果也能佐证:
-- 1、 T1表,k3索引在前面
PRIMARY KEY (`c1`),
UNIQUE KEY `k3` (`c3`),
UNIQUE KEY `k2` (`c2`)
T@2: | | | | | | | | opt: (null): starting struct
T@2: | | | | | | | | opt: table: "`t1`"
T@2: | | | | | | | | opt: field: "c3" (C3在前面,因此最后使用k3)
T@2: | | | | | | | | >convert_string
T@2: | | | | | | | | | >alloc_root
T@2: | | | | | | | | | | enter: root: 0x40a8068
T@2: | | | | | | | | | | exit: ptr: 0x4b41ab0
T@2: | | | | | | | | | T@2: | | | | | | | | T@2: | | | | | | | | opt: equals: "'Louise Garrett'"
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: (null): ending struct
T@2: | | | | | | | | opt: Key_use: optimize= 0 used_tables=0x0 ref_table_rows= 18446744073709551615 keypart_map= 1
T@2: | | | | | | | | opt: (null): starting struct
T@2: | | | | | | | | opt: table: "`t1`"
T@2: | | | | | | | | opt: field: "c2"
T@2: | | | | | | | | opt: equals: "22896242"
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: (null): ending struct
T@2: | | | | | | | | opt: Key_use: optimize= 0 used_tables=0x0 ref_table_rows= 18446744073709551615 keypart_map= 1
T@2: | | | | | | | | opt: (null): starting struct
T@2: | | | | | | | | opt: table: "`t1`"
T@2: | | | | | | | | opt: field: "c2"
T@2: | | | | | | | | opt: equals: "22896242"
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: (null): ending struct
T@2: | | | | | | | | opt: ref_optimizer_key_uses: ending struct
T@2: | | | | | | | | opt: (null): ending struct
-- 2、 T2表,k2索引在前面
PRIMARY KEY (`c1`),
UNIQUE KEY `k2` (`c2`),
UNIQUE KEY `k3` (`c3`)
T@2: | | | | | | | | opt: (null): starting struct
T@2: | | | | | | | | opt: table: "`t2`"
T@2: | | | | | | | | opt: field: "c2" (C2在前面因此使用k2索引)
T@2: | | | | | | | | opt: equals: "22896242"
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: (null): ending struct
T@2: | | | | | | | | opt: Key_use: optimize= 0 used_tables=0x0 ref_table_rows= 18446744073709551615 keypart_map= 1
T@2: | | | | | | | | opt: (null): starting struct
T@2: | | | | | | | | opt: table: "`t2`"
T@2: | | | | | | | | opt: field: "c3"
T@2: | | | | | | | | >convert_string
T@2: | | | | | | | | | >alloc_root
T@2: | | | | | | | | | | enter: root: 0x40a8068
T@2: | | | | | | | | | | exit: ptr: 0x4b41ab0
T@2: | | | | | | | | | T@2: | | | | | | | | T@2: | | | | | | | | opt: equals: "'Louise Garrett'"
T@2: | | | | | | | | opt: null_rejecting: 0
T@2: | | | | | | | | opt: (null): ending struct
T@2: | | | | | | | | opt: ref_optimizer_key_uses: ending struct
T@2: | | | | | | | | opt: (null): ending struct
4. 问题延伸
到这里,我们不禁有疑问,这两个索引的代价真的是一样吗?
就让我们用 mysqlslap
来做个简单对比测试吧:
-- 测试1:对c2列随机point select
mysqlslap -hlocalhost -uroot -Smysql.sock --no-drop --create-schema X -i 3 --number-of-queries 1000000 -q "set @xid = cast(round(rand()*2147265929) as unsigned); select * from t1 where c2 = @xid" -c 8
...
Average number of seconds to run all queries: 9.483 seconds
...
-- 测试2:对c3列随机point select
mysqlslap -hlocalhost -uroot -Smysql.sock --no-drop --create-schema X -i 3 --number-of-queries 1000000 -q "set @xid = concat('u',cast(round(rand()*2147265929) as unsigned)); select * from t1 where c3 = @xid" -c 8
...
Average number of seconds to run all queries: 10.360 seconds
...
可以看到,如果是走 c3 列索引,耗时会比走 c2 列索引多出来约 7% ~ 9%(在我的环境下测试的结果,不同环境、不同数据量可能也不同)。
看来,MySQL优化器还是有必要进一步提高的哟 :)
测试使用版本:GreatSQL 8.0.25
(MySQL 5.6.39结果亦是如此)。
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