我正在使用 MySQL 5.5.37。目前无法升级。我有这张 table
CREATE TABLE `my_classroom` (
`ID` varchar(32) COLLATE utf8_bin NOT NULL DEFAULT '',
`CLASSROOM_NAME` varchar(100) COLLATE utf8_bin NOT NULL,
`ACCESS_CODE_ID` varchar(32) COLLATE utf8_bin DEFAULT NULL,
`TEACHER_ACCESS_CODE_ID` varchar(32) COLLATE utf8_bin DEFAULT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `UK_my_classroom` (`ACCESS_CODE_ID`),
UNIQUE KEY `UK2_my_classroom` (`TEACHER_ACCESS_CODE_ID`),
KEY `FK2_my_classroom` (`CLASSROOM_SCHEDULE_ID`),
CONSTRAINT `FK3_my_classroom` FOREIGN KEY (`TEACHER_ACCESS_CODE_ID`) REFERENCES `my_reg_code` (`ID`) ON UPDATE NO ACTION,
CONSTRAINT `FK_my_classroom` FOREIGN KEY (`ACCESS_CODE_ID`) REFERENCES `my_reg_code` (`ID`) ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin
注意 ACCESS_CODE_ID 和 TEACHER_ACCESS_CODE_ID 列上的索引。然而,当这个查询运行时(它是由 Hibernate 生成的,这就是它看起来有点古怪的原因),请注意正在发生的全表扫描......
mysql> explain select davesclass0_.id as id1_7_, davesclass0_.ACCESS_CODE_ID as ACCESS_13_7_,
davesclass0_.CLASSROOM_NAME as CLASSROO7_7_, davesclass0_.TEACHER_ACCESS_CODE_ID as TEACHER15_7_
from my_classroom davesclass0_
left outer join my_reg_code myregcode1_ on davesclass0_.ACCESS_CODE_ID=myregcode1_.ID
left outer join my_reg_code accesscode2_ on davesclass0_.TEACHER_ACCESS_CODE_ID=accesscode2_.ID
where myregcode1_.ACCESS_CODE='ABCDEF' or accesscode2_.ACCESS_CODE='ABCDEF';
+----+-------------+--------------+--------+---------------+---------+---------+-------------------------------------------+---------+-------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+-------------+--------------+--------+---------------+---------+---------+-------------------------------------------+---------+-------------+
| 1 | SIMPLE | davesclass0_ | ALL | NULL | NULL | NULL | NULL | 1914867 | |
| 1 | SIMPLE | myregcode1_ | eq_ref | PRIMARY | PRIMARY | 98 | my_db.davesclass0_.ACCESS_CODE_ID | 1 | |
| 1 | SIMPLE | accesscode2_ | eq_ref | PRIMARY | PRIMARY | 98 | my_db.davesclass0_.TEACHER_ACCESS_CODE_ID | 1 | Using where |
+----+-------------+--------------+--------+---------------+---------+---------+-------------------------------------------+---------+-------------+
有什么方法可以重写它以返回相同的结果,但让 MySQL 理解使用 my_classroom 表上的索引?
编辑: 针对lsemi的建议,MySql的explain plan ...
mysql> explain select davesclass0_.id as id1_7_, davesclass0_.ACCESS_CODE_ID as ACCESS_13_7_, davesclass0_.TEACHER_ACCESS_CODE_ID as TEACHER15_7_ from my_classroom davesclass0_ where davesclass0_.ACCESS_CODE_ID in (select myregcode1_.ID from my_reg_code myregcode1_ where myregcode1_.ACCESS_CODE='ABCDEF') or davesclass0_.TEACHER_ACCESS_CODE_ID in (select myregcode2_.ID from my_reg_code myregcode2_ where myregcode2_.ACCESS_CODE='ABCDEF');
+----+--------------------+--------------+------+---------------+------+---------+------+--------+-----------------------------------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+--------------+------+---------------+------+---------+------+--------+-----------------------------------------------------+
| 1 | PRIMARY | davesclass0_ | ALL | NULL | NULL | NULL | NULL | 216280 | Using where |
| 3 | DEPENDENT SUBQUERY | NULL | NULL | NULL | NULL | NULL | NULL | NULL | Impossible WHERE noticed after reading const tables |
| 2 | DEPENDENT SUBQUERY | NULL | NULL | NULL | NULL | NULL | NULL | NULL | Impossible WHERE noticed after reading const tables |
+----+--------------------+--------------+------+---------------+------+---------+------+--------+-----------------------------------------------------+
编辑 2: 解释计划
mysql> explain select davesclass0_.id as id1_7_, davesclass0_.ACCESS_CODE_ID as ACCESS_13_7_, davesclass0_.TEACHER_ACCESS_CODE_ID as TEACHER15_7_ from mY_classroom davesclass0_ where davesclass0_.ACCESS_CODE_ID in (select myregcode1_.ID from my_reg_code myregcode1_ where myregcode1_.ACCESS_CODE='0008F0');
+----+--------------------+--------------+-------+---------------------------+-------------------+---------+-------+--------+-------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+--------------------+--------------+-------+---------------------------+-------------------+---------+-------+--------+-------------+
| 1 | PRIMARY | davesclass0_ | ALL | NULL | NULL | NULL | NULL | 216280 | Using where |
| 2 | DEPENDENT SUBQUERY | myregcode1_ | const | PRIMARY,UK_my_reg_code | UK_my_reg_code | 98 | const | 1 | Using index |
+----+--------------------+--------------+-------+---------------------------+-------------------+---------+-------+--------+-------------+
2 rows in set (0.00 sec)
mysql> explain select davesclass0_.id as id1_7_, davesclass0_.ACCESS_CODE_ID as ACCESS_13_7_, davesclass0_.TEACHER_ACCESS_CODE_ID as TEACHER15_7_ from mY_classroom davesclass0_ where davesclass0_.ACCESS_CODE_ID = 'ABCEF';
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
| id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
| 1 | SIMPLE | NULL | NULL | NULL | NULL | NULL | NULL | NULL | Impossible WHERE noticed after reading const tables |
+----+-------------+-------+------+---------------+------+---------+------+------+-----------------------------------------------------+
最佳答案
注意:我假设您只需要那些匹配的记录并且这个数字很小。否则,您将阅读所有内容,因此索引是无用的。因此,MySQL 不使用它们是正确的。
根据这个假设,我将开始将这些 JOIN 重写为直接 JOIN(而不是 LEFT JOIN),并验证是否存在一个索引,例如
CREATE INDEX my_reg_code_ndx ON my_reg_code(ACCESS_CODE);
然后您也许可以使用单个 JOIN(如果单个记录同时将访问权限和教师访问权限设置为 ABCDEF 的 ID,结果可能会略有不同):
JOIN my_reg_code AS mrc ON (mrc.ID = ACCESS_CODE_ID OR mrc.ID = TEACHER_ACCESS_CODE_ID)
或者您也可以将 WHERE 重写为完全不同的查询:
SELECT ... FROM my_classroom AS mc
WHERE mc.ACCESS_CODE_ID IN
(SELECT ID from my_reg_code WHERE ACCESS_CODE='ABCDEF')
OR mc.TEACHER_ACCESS_CODE_ID IN (...)
通常 MySQL 在执行每个查询之前 会运行一次计算,并且此计算基于查询结构和约束。这意味着在某些情况下,MySQL 无法在分配给初始查询研究的几毫秒内进行计算。
此信息的一个来源是更新索引统计信息的 ANALYZE 命令。
但这可能还不够,因为如果你做这样的查询
SELECT * FROM table WHERE indexedfield = value
可以根据值给出字段基数的一些估计;如果你这样做
SELECT * FROM table WHERE indexfield = FUNCTION(value)
或
SELECT * FROM table WHERE indexfield = SELECT(...)
那么 MySQL 可能不会运行该函数或查询,因此无法实际执行外部 SELECT 的分析。当实际运行查询时,函数或 SELECT 将转换为一组值,然后MySQL 将运行外部查询分析并使用索引,或者不使用索引。
因此,一个复杂的查询(不是基于 JOIN 的)可能会使用您期望的索引、您不期望的索引,或者根本没有。
输入提示
您可以建议 MySQL to use a specific index :
SELECT * FROM my_classroom AS mc USE INDEX (indexToUse)
WHERE mc.ACCESS_CODE_ID IN (...) ...
您也可以使用FORCE 代替USE。
关于mysql - 为什么 MySQL 在运行查询时不使用我的索引?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39904206/