解决like‘% str %’索引不被使用的方法

问题：解决like‘%字符串%’时索引不被使用的方法
解决方法：覆盖索引
新建表tbl_user

CREATE TABLE `tbl_user` (
  `id` int NOT NULL,
  `name` varchar(20) DEFAULT NULL,
  `age` int DEFAULT NULL,
  `email` varchar(20) DEFAULT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci;

插入字段

INSERT INTO `test`.`tbl_user`(`id`, `name`, `age`, `email`) VALUES (1, '1aa1', 21, '[email protected]');
INSERT INTO `test`.`tbl_user`(`id`, `name`, `age`, `email`) VALUES (2, '2aa2', 222, '[email protected]');
INSERT INTO `test`.`tbl_user`(`id`, `name`, `age`, `email`) VALUES (3, '3aa3', 265, '[email protected]');
INSERT INTO `test`.`tbl_user`(`id`, `name`, `age`, `email`) VALUES (4, '4aa4', 21, '[email protected]');

 SELECT name,age FROM tbl_user WHERE name like '%aa%';

SELECT name,age FROM tbl_user WHERE name like 'aa%';

SELECT name,age FROM tbl_user WHERE name like '%aa';

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 EXPLAIN SELECT name,age FROM tbl_user WHERE name like '%aa%';

#创建索引

CREATE INDEX idx_user_nameAge ON tbl_user(name,age);

创建索引之后再查一次

EXPLAIN SELECT name,age FROM tbl_user WHERE name like '%aa%';

EXPLAIN SELECT id FROM tbl_user WHERE name like '%aa%';

 EXPLAIN SELECT name FROM tbl_user WHERE name like '%aa%';

 EXPLAIN SELECT age FROM tbl_user WHERE name like '%aa%';

EXPLAIN SELECT name,age FROM tbl_user WHERE name like '%aa%';
	 EXPLAIN SELECT id,age FROM tbl_user WHERE name like '%aa%';
	  EXPLAIN SELECT id,name FROM tbl_user WHERE name like '%aa%';

索引都没有失效

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EXPLAIN SELECT * FROM tbl_user WHERE name like '%aa%';

使用 * 时索引失效了

EXPLAIN SELECT name,age,email FROM tbl_user WHERE name like '%aa%';