# 1、in
select * from emp where age=18 or age=38 or age=28;
select * from emp where age in (18,38,28);
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# 子查询的思路
select * from emp where dep_id in
(select id from dep where name="技术" or name="销售");
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# 链表的思路
select * from emp inner join dep
on emp.dep_id = dep.id
where dep.name in ("技术","销售");
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# not in不支持null
mysql> select * from dep;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+
4 rows in set (0.00 sec)
mysql> insert into emp values(7,‘lili‘,‘female‘,48,null);
Query OK, 1 row affected (0.03 sec)
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mysql> select * from emp
-> ;
+----+------------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+------------+--------+------+--------+
| 1 | egon | male | 18 | 200 |
| 2 | alex | female | 48 | 201 |
| 3 | wupeiqi | male | 38 | 201 |
| 4 | yuanhao | female | 28 | 202 |
| 5 | liwenzhou | male | 18 | 200 |
| 6 | jingliyang | female | 18 | 204 |
| 7 | lili | female | 48 | NULL |
+----+------------+--------+------+--------+
7 rows in set (0.00 sec)
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mysql>
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查询出有员工的部门，
select * from dep where id in
(select distinct dep_id from emp);
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查询出没有员工的部门，
select * from dep where id not in
(select distinct dep_id from emp);
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select * from dep where id not in
(select distinct dep_id from emp where dep_id is not null);
2.子查询any和all
# any后也跟子查询语句，与in不一样的地方在哪里
# in (子查询语句)
# in (值1,值2,值3)
# 而any只能跟子查询语句
# any必须跟比较运算符配合使用
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select * from emp where dep_id in
(select id from dep where name in ("技术","人力资源"));
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select * from emp where dep_id = any
(select id from dep where name in ("技术","人力资源"));
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select * from emp where dep_id not in
(select id from dep where name in ("技术","人力资源"));
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select * from emp where ! (dep_id = any(select id from dep where name in ("技术","人力资源")));
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查询出那些薪资比所有部门的平均薪资都高的员工=》薪资在所有部门平均线以上的狗币资本家
select * from employee where salary > all
(select avg(salary) from employee where depart_id is not null group by depart_id);
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查询出那些薪资比所有部门的平均薪资都低的员工=》薪资在所有部门平均线以下的无产阶级劳苦大众
select * from employee where salary < all
(select avg(salary) from employee where depart_id is not null group by depart_id);
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查询出那些薪资比任意一个部门的平均薪资高的员工=》薪资在任一部门平均线以上的员工
select * from employee where salary > any
(select avg(salary) from employee where depart_id is not null group by depart_id);
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select * from employee where salary < any
(select avg(salary) from employee where depart_id is not null group by depart_id);
3.子查询exists
# 准备数据
create database db13;
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use db13
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create table student(
id int primary key auto_increment,
name varchar(16)
);
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create table course(
id int primary key auto_increment,
name varchar(16),
comment varchar(20)
);
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create table student2course(
id int primary key auto_increment,
sid int,
cid int,
foreign key(sid) references student(id),
foreign key(cid) references course(id)
);
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insert into student(name) values
("egon"),
("lili"),
("jack"),
("tom");
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insert into course(name,comment) values
("数据库","数据仓库"),
("数学","根本学不会"),
("英语","鸟语花香");
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insert into student2course(sid,cid) values
(1,1),
(1,2),
(1,3),
(2,1),
(2,2),
(3,2);
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准备数据
# exists vs in
# in的效果 高于 exists
# 见博客：https://www.cnblogs.com/linhaifeng/articles/7267596.html#_label4
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select * from 表1 where exists (select * from 表2);
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# 例如：查询有员工的部门=》
select * from dep where exists (select * from emp where dep.id=emp.dep_id);
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# not exists的效果 高于 not in
select * from dep where not exists (select * from emp where 203=emp.dep_id);
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# 例：查询选修了所有课程的学生id、name:
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# 实现方式一：选修了三门课程的学生就是选修了所有课程的学生
select s.id,s.name from student as s inner join student2course as sc
on s.id = sc.sid
group by sc.sid
having count(sc.cid) = (select count(id) from course);
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# 实现方式二：找到这样的学生，该学生不存在没有选修过的课程
select * from student as s where not exists (
select * from course as c where not exists (
select * from student2course as sc where sc.sid = s.id and sc.cid = c.id
)
);
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select * from student as s where not exists (
select * from course as c where not exists (
select * from student2course as sc where sc.sid = s.id and sc.cid = c.id
)
);
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学生记录可以过滤出来，一定是子查询内没有记录
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for 学生: # s.id=2
for 课程: # c.id=1
for 学生2课程: # sc.sid = 2 and sc.cid = 1
pass
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==================================
for sid in [1,2,3,4]:
for cid in [1,2,3]:
(sid,cid)
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最外层循环一次
# (1,1)
# (1,2)
# (1,3)
最外层循环二次
# (2,1)
# (2,2)
# (2,3)
最外层循环三次
# (3,1)
# (3,2)
# (3,3)
最外层循环四次
# (4,1)
# (4,2)
# (4,3)
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===================================
# 例2、查询没有选择所有课程的学生，即没有全选的学生。=》找出这样的学生，存在没有选修过的课程
select * from student as s where exists (
select * from course as c where not exists (
select * from student2course as sc where sc.sid = s.id and sc.cid = c.id
)
);
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# 例3、查询一门课也没有选的学生=》找出这样的学生，不存在选修过的课程