MySQL索引失效实践

导入sql

CREATE TABLE `employees` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `name` varchar(24) NOT NULL DEFAULT '' COMMENT '姓名',
  `age` int(11) NOT NULL DEFAULT '0' COMMENT '年龄',
  `position` varchar(20) NOT NULL DEFAULT '' COMMENT '职位',
  `hire_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP COMMENT '入职时间',
  PRIMARY KEY (`id`),
  KEY `idx_name_age_position` (`name`,`age`,`position`) USING BTREE
) ENGINE=InnoDB AUTO_INCREMENT=4 DEFAULT CHARSET=utf8 COMMENT='员工记录表';

INSERT INTO employees(name,age,position,hire_time) VALUES('LiLei',22,'manager',NOW());
INSERT INTO employees(name,age,position,hire_time) VALUES('HanMeimei', 23,'dev',NOW());
INSERT INTO employees(name,age,position,hire_time) VALUES('Lucy',23,'dev',NOW());

最佳实践

1. 全值匹配

EXPLAIN SELECT * FROM employees WHERE name= 'LiLei';
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EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 22;
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EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 22 AND position ='manager';
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2.最佳左前缀法则

如果索引了多列,要遵守最左前缀法则。指的是查询从索引的最左前列开始并且不跳过索引中的列。
EXPLAIN SELECT * FROM employees WHERE age = 22 AND position ='manager';
EXPLAIN SELECT * FROM employees WHERE position = 'manager';
EXPLAIN SELECT * FROM employees WHERE name = 'LiLei';

3.不在索引列上做任何操作(计算、函数、(自动or手动)类型转换),会导致索引失效而转向全表扫描

EXPLAIN SELECT * FROM employees WHERE name = 'LiLei';
EXPLAIN SELECT * FROM employees WHERE left(name,3) = 'LiLei';
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4.存储引擎不能使用索引中范围条件右边的列

EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 22 AND position ='manager';
EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age > 22 AND position ='manager';
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5.尽量使用覆盖索引(只访问索引的查询(索引列包含查询列)),减少select *语句

EXPLAIN SELECT NAME,age FROM employees WHERE NAME= 'LiLei' AND age = 23 AND POSITION ='manager';
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EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 23 AND position ='manager';
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6.mysql在使用不等于(!=或者<>)的时候无法使用索引会导致全表扫描

EXPLAIN SELECT * FROM employees WHERE name != 'LiLei'
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7.is null,is not null 也无法使用索引

EXPLAIN SELECT * FROM employees WHERE name is null
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8.like以通配符开头(’$abc…’)mysql索引失效会变成全表扫描操作

EXPLAIN SELECT * FROM employees WHERE name like '%Lei'
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EXPLAIN SELECT * FROM employees WHERE NAME LIKE 'Lei%'
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问题:解决like’%字符串%'索引不被使用的方法?

a)使用覆盖索引,查询字段必须是建立覆盖索引字段
EXPLAIN SELECT name,age,position FROM employees WHERE name like '%Lei%';
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b)当覆盖索引指向的字段是varchar(380)及380以上的字段时,覆盖索引会失效!

9.字符串不加单引号索引失效

EXPLAIN SELECT * FROM employees WHERE name = '1000';
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EXPLAIN SELECT * FROM employees WHERE name = 1000;
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10.少用or,用它连接时很多情况下索引会失效

EXPLAIN SELECT * FROM employees WHERE name = 'LiLei' or name = 'HanMeimei';
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转载自blog.csdn.net/weixin_44374871/article/details/114746948