Mysql 中 exists 和 in 的区别_mysql中 in 和 exists 区别

最近刷到面试题：Mysql 中 exists 和 in 的区别，先说下答案。

 

下面将主查询的表称为外表；子查询的表称为内表。exists 与 in 的区别如下：

• 子查询使用 exists，会先进行主查询，将查询到的每行数据循环带入子查询校验是否存在，过滤出整体的返回数据；子查询使用 in，会先进行子查询获取结果集，然后主查询匹配子查询的结果集，返回数据
• 外表内表相对大小情况不一样时，查询效率不一样：两表大小相当，in 和 exists 差别不大；内表大，用 exists 效率较高；内表小，用 in 效率较高。
• 不管外表与内表的大小，not exists 的效率一般要高于 not in，跟子查询的索引访问类型有关。

 

建表、造数据，验证一下以上答案。

# 建表 student1
drop table if exists student1;
create table student1(
sid int primary key auto_increment,
sname varchar(40)
);
 
# 建存储过程给表 student1，插入1000条数据
drop procedure if exists addStudent1;
create procedure addStudent1()
BEGIN
	declare idx int;
	set idx = 1;
	while idx <= 1000 DO
	insert into student1 values(null, concat('student-', idx));
	set idx = idx + 1;
  end while;
end;
 
call addStudent1();
 
select * from student1;
 
# 建表 student2
drop table if exists student2;
create table student2(
sid int primary key auto_increment,
sname varchar(40)
);
 
# 建存储过程给表 student2，插入100000条数据
drop procedure if exists addStudent2;
create procedure addStudent2()
BEGIN
	declare idx int;
	set idx = 1;
	while idx <= 100000 DO
	insert into student2 values(null, concat('student-', idx));
	set idx = idx + 1;
  end while;
end;
 
call addStudent2();
 
select * from student2;

 

in 与 exists 的查询 SQL

select count(1) from student1 where sname in (select sname from student2);
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
select count(1) from student2 where sname in (select sname from student1);
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);

执行时间：

[SQL] select count(1) from student1 where sname in (select sname from student2);
受影响的行: 0
时间: 0.092s
 
[SQL] 
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.076s
 
[SQL] 
select count(1) from student2 where sname in (select sname from student1);
受影响的行: 0
时间: 14.820s
 
[SQL] 
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);
受影响的行: 0
时间: 15.144s

结论：student2 大表在内适用 exists，所以第 2 条 SQL 比第 1 条快；student1 小表在内适用 in，所以第 3 条 SQL 比第 4 条快。

 

not in 与 not exists 的查询 SQL

select count(1) from student1 where sname not in (select sname from student2);
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
select count(1) from student2 where sname not in (select sname from student1);
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);

执行时间：

[SQL] 
select count(1) from student1 where sname not in (select sname from student2);
受影响的行: 0
时间: 0.079s
 
[SQL] 
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.075s
 
[SQL] select count(1) from student2 where sname not in (select sname from student1);
受影响的行: 0
时间: 15.797s
 
[SQL] 
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
受影响的行: 0
时间: 15.160s

结论：not exists 性能高于 not in

 

给 student1、student2 sname 字段，加上索引，上述结论仍然成立。

create index idx_1 on student1(sname);
create index idx_2 on student2(sname);

执行时间：

[SQL] select count(1) from student1 where sname in (select sname from student2);
受影响的行: 0
时间: 0.022s
 
[SQL] 
select count(1) from student1 where exists (select sname from student2 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.014s
 
[SQL] 
select count(1) from student2 where sname in (select sname from student1);
受影响的行: 0
时间: 0.379s
 
[SQL] 
select count(1) from student2 where exists (select sname from student1 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.373s
 
[SQL] 
select count(1) from student1 where sname not in (select sname from student2);
受影响的行: 0
时间: 0.006s
 
[SQL] 
select count(1) from student1 where not exists (select sname from student2 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.006s
 
[SQL] 
select count(1) from student2 where sname not in (select sname from student1);
受影响的行: 0
时间: 0.455s
 
[SQL] 
select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
受影响的行: 0
时间: 0.418s

 

再细看一下，not in 与 not exists 查询索引使用情况

not in，子查询使用了 index_subquery 访问类型

EXPLAIN EXTENDED select count(1) from student2 where sname not in (select sname from student1);
SHOW WARNINGS;

 

not exists，子查询使用了 ref 访问类型

EXPLAIN EXTENDED select count(1) from student2 where not exists (select sname from student1 where student2.sname = student1.sname);
SHOW WARNINGS;

 

 

---

【Java面试题与答案】整理推荐

• 基础与语法
• 集合
• 网络编程
• 并发编程
• Web
• 安全
• 设计模式
• 框架
• 算法与数据结构
• 异常
• 文件解析与生成
• Linux
• MySQL
• Oracle
• Redis
• Dubbo