索引最佳实践
使用的表
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';
 

EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 22;
 

EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 22 AND position ='manager';
 

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';
 

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';
 

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';
 
EXPLAIN SELECT * FROM employees WHERE name= 'LiLei' AND age = 23 AND position ='manager';
 

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'
 

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

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'
 
EXPLAIN SELECT * FROM employees WHERE name like 'Lei%'
 

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%';
 
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';
EXPLAIN SELECT * FROM employees WHERE name = 1000;
 

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';
 

总结：
 
like KK%相当于=常量，%KK和%KK% 相当于范围
————————————————
版权声明：本文为CSDN博主「王林冲」的原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接及本声明。
原文链接：https://blog.csdn.net/qq_38138069/article/details/82998658