多表查询 – Egon林海峰

一介绍

本节主题

多表连接查询
复合条件连接查询
子查询

准备表

表department与employee

#建表
create table department(
id int,
name varchar(20) 
);

create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum('male','female') not null default 'male',
age int,
dep_id int
);

#插入数据
insert into department values
(200,'技术'),
(201,'人力资源'),
(202,'销售'),
(203,'运营');

insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
('alex','female',48,201),
('wupeiqi','male',38,201),
('yuanhao','female',28,202),
('liwenzhou','male',18,200),
('jingliyang','female',18,204)
;


#查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+

mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

mysql> select * from employee;
+----+------------+--------+------+--------+
| 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 |
+----+------------+--------+------+--------+

二多表连接查询

#重点：外链接语法

SELECT 字段列表
    FROM 表1 INNER|LEFT|RIGHT JOIN 表2
    ON 表1.字段 = 表2.字段;

1 交叉连接：不适用任何匹配条件。生成笛卡尔积

mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name       | sex    | age  | dep_id | id   | name         |
+----+------------+--------+------+--------+------+--------------+
|  1 | egon       | male   |   18 |    200 |  200 | 技术         |
|  1 | egon       | male   |   18 |    200 |  201 | 人力资源     |
|  1 | egon       | male   |   18 |    200 |  202 | 销售         |
|  1 | egon       | male   |   18 |    200 |  203 | 运营         |
|  2 | alex       | female |   48 |    201 |  200 | 技术         |
|  2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|  2 | alex       | female |   48 |    201 |  202 | 销售         |
|  2 | alex       | female |   48 |    201 |  203 | 运营         |
|  3 | wupeiqi    | male   |   38 |    201 |  200 | 技术         |
|  3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|  3 | wupeiqi    | male   |   38 |    201 |  202 | 销售         |
|  3 | wupeiqi    | male   |   38 |    201 |  203 | 运营         |
|  4 | yuanhao    | female |   28 |    202 |  200 | 技术         |
|  4 | yuanhao    | female |   28 |    202 |  201 | 人力资源     |
|  4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|  4 | yuanhao    | female |   28 |    202 |  203 | 运营         |
|  5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
+----+------------+--------+------+--------+------+--------------+

2 内连接：只连接匹配的行

#找两张表共有的部分，相当于利用条件从笛卡尔积结果中筛选出了正确的结果
#department没有204这个部门，因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
+----+-----------+------+--------+--------------+
| id | name      | age  | sex    | name         |
+----+-----------+------+--------+--------------+
|  1 | egon      |   18 | male   | 技术         |
|  2 | alex      |   48 | female | 人力资源     |
|  3 | wupeiqi   |   38 | male   | 人力资源     |
|  4 | yuanhao   |   28 | female | 销售         |
|  5 | liwenzhou |   18 | male   | 技术         |
+----+-----------+------+--------+--------------+

#上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;

3 外链接之左连接：优先显示左表全部记录

#以左表为准，即找出所有员工信息，当然包括没有部门的员工
#本质就是：在内连接的基础上增加左边有右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name       | depart_name  |
+----+------------+--------------+
|  1 | egon       | 技术         |
|  5 | liwenzhou  | 技术         |
|  2 | alex       | 人力资源     |
|  3 | wupeiqi    | 人力资源     |
|  4 | yuanhao    | 销售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+

4 外链接之右连接：优先显示右表全部记录

#以右表为准，即找出所有部门信息，包括没有员工的部门
#本质就是：在内连接的基础上增加右边有左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id   | name      | depart_name  |
+------+-----------+--------------+
|    1 | egon      | 技术         |
|    2 | alex      | 人力资源     |
|    3 | wupeiqi   | 人力资源     |
|    4 | yuanhao   | 销售         |
|    5 | liwenzhou | 技术         |
| NULL | NULL      | 运营         |
+------+-----------+--------------+

5 全外连接：显示左右两个表全部记录

全外连接：在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意：mysql不支持全外连接 full JOIN
#强调：mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|    2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|    3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|    4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|    6 | jingliyang | female |   18 |    204 | NULL | NULL         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+

#注意 union与union all的区别：union会去掉相同的纪录

三符合条件连接查询

#示例1：以内连接的方式查询employee和department表，并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department
    on employee.dep_id = department.id
    where age > 25;

#示例2：以内连接的方式查询employee和department表，并且以age字段的升序方式显示
select employee.id,employee.name,employee.age,department.name from employee,department
    where employee.dep_id = department.id
    and age > 25
    order by age asc;

四子查询

#1：子查询是将一个查询语句嵌套在另一个查询语句中。
#2：内层查询语句的查询结果，可以为外层查询语句提供查询条件。
#3：子查询中可以包含：IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4：还可以包含比较运算符：= 、 !=、> 、<等

1 带IN关键字的子查询

#查询平均年龄在25岁以上的部门名
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);

#查看技术部员工姓名
select name from employee
    where dep_id in 
        (select id from department where name='技术');

#查看不足1人的部门名(子查询得到的是有人的部门id)
select name from department where id not in (select distinct dep_id from employee);

！！！注意not in

not in 无法处理null的值，即子查询中如果存在null的值，not in将无法处理，如下

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 | xxx | male | 19 | NULL |
+----+------------+--------+------+--------+
7 rows in set (0.00 sec)

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)

# 子查询中存在null
mysql> select * from dep where id not in (select distinct dep_id from emp);
Empty set (0.00 sec)

# 解决方案如下
mysql> select * from dep where id not in (select distinct dep_id from emp where dep_id is not null);
+------+--------+
| id | name |
+------+--------+
| 203 | 运营 |
+------+--------+
1 row in set (0.00 sec)

mysql>

2 带ANY关键字的子查询

#在 SQL 中 ANY 和 SOME 是同义词，SOME 的用法和功能和 ANY 一模一样。

# ANY 和 IN 运算符不同之处1
ANY 必须和其他的比较运算符共同使用，而且ANY必须将比较运算符放在 ANY 关键字之前，所比较的值需要匹配子查询中的任意一个值，这也就是 ANY 在英文中所表示的意义

例如：使用 IN 和使用 ANY运算符得到的结果是一致的
select * from employee where salary = any (
select max(salary) from employee group by depart_id);

select * from employee where salary in (
select max(salary) from employee group by depart_id);

结论：也就是说“=ANY”等价于 IN 运算符，而“<>ANY”则等价于 NOT IN 运算符

# ANY和 IN 运算符不同之处2
ANY 运算符不能与固定的集合相匹配，比如下面的 SQL 语句是错误的

SELECT
*
FROM
T_Book
WHERE
FYearPublished < ANY (2001, 2003, 2005)

3 带ALL关键字的子查询

# all同any类似，只不过all表示的是所有，any表示任一

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 group by depart_id);

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 group by depart_id);





查询出那些薪资比任意一个部门的平均薪资低的员工=》薪资在任一部门平均线以下的员工
select * from employee where salary < any ( select avg(salary) from employee group by depart_id);

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 group by depart_id);

4 带比较运算符的子查询

#比较运算符：=、!=、>、>=、<、<=、<>
#查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from emp where age > (select avg(age) from emp);
+---------+------+
| name | age |
+---------+------+
| alex | 48 |
| wupeiqi | 38 |
+---------+------+
2 rows in set (0.00 sec)


#查询大于部门内平均年龄的员工名、年龄
select t1.name,t1.age from emp t1
inner join 
(select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;

5 带EXISTS关键字的子查询

EXISTS关字键字表示存在。在使用EXISTS关键字时，内层查询语句不返回查询的记录。
而是返回一个真假值。True或False
当返回True时，外层查询语句将进行查询；当返回值为False时，外层查询语句不进行查询

#department表中存在dept_id=203，Ture
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=200);
+----+------------+--------+------+--------+
| 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 |
+----+------------+--------+------+--------+

#department表中存在dept_id=205，False
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=204);
Empty set (0.00 sec)

5.1 in与exists

！！！！！！当in和exists在查询效率上比较时，in查询的效率快于exists的查询效率！！！！！！

==============================exists==============================
# exists
exists后面一般都是子查询，后面的子查询被称做相关子查询（即与主语句相关），当子查询返回行数时，exists条件返回true，
否则返回false，exists是不返回列表的值的，exists只在乎括号里的数据能不能查找出来，是否存在这样的记录。

# 例
查询出那些班级里有学生的班级
select * from class where exists (select * from stu where stu.cid=class.id)

# exists的执行原理为：
1、依次执行外部查询：即select * from class 
2、然后为外部查询返回的每一行分别执行一次子查询：即(select * from stu where stu.cid=class.cid)
3、子查询如果返回行，则exists条件成立，条件成立则输出外部查询取出的那条记录

==============================in==============================
# in
in后跟的都是子查询，in()后面的子查询 是返回结果集的

# 例
查询和所有女生年龄相同的男生
select * from stu where sex='男' and age in(select age from stu where sex='女')

# in的执行原理为：
in()的执行次序和exists()不一样,in()的子查询会先产生结果集,
然后主查询再去结果集里去找符合要求的字段列表去.符合要求的输出,反之则不输出.

5.2 not in与 not exists

!!!!!!not exists查询的效率远远高与not in查询的效率。!!!!!!

==============================not in==============================
not in()子查询的执行顺序是：
为了证明not in成立，即找不到，需要一条一条地查询表，符合要求才返回子查询的结果集，不符合的就继续查询下一条记录，直到把表中的记录查询完，只能查询全部记录才能证明，并没有用到索引。
                
==============================not exists==============================
not exists：
如果主查询表中记录少，子查询表中记录多，并有索引。
例如:查询那些班级中没有学生的班级
select * from class

where not exists

(select * from student where student.cid = class.cid)

not exists的执行顺序是：
在表中查询，是根据索引查询的，如果存在就返回true，如果不存在就返回false，不会每条记录都去查询。

应用-准备数据

create database db13;

use db13

create table student(
    id int primary key auto_increment,
    name varchar(16)
);

create table course(
    id int primary key auto_increment,
    name varchar(16),
    comment varchar(20)
);

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)
);


insert into student(name) values
("egon"),
("lili"),
("jack"),
("tom");

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");


insert into student2course(sid,cid) values
(1,1),
(1,2),
(1,3),
(2,1),
(2,2),
(3,2);

示例

# 1、查询选修了所有课程的学生id、name:（即该学生根本就不存在一门他没有选的课程。）
select * from student s where not exists
    (select * from course c where not exists
        (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));


select s.name from student as s
inner join student2course as sc
on s.id=sc.sid
group by s.name 
having count(sc.id) = (select count(id) from course);

# 2、查询没有选择所有课程的学生，即没有全选的学生。（存在这样的一个学生，他至少有一门课没有选）
select * from student s where exists
    (select * from course c where not exists
        (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));

# 3、查询一门课也没有选的学生。（不存这样的一个学生，他至少选修一门课程）
select * from student s where not exists
    (select * from course c where exists
        (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));

# 4、查询至少选修了一门课程的学生。
select * from student s where exists
    (select * from course c where exists
        (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));

练习：查询每个部门最新入职的那位员工

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一 介绍

二 多表连接查询

三 符合条件连接查询

四 子查询

练习：查询每个部门最新入职的那位员工

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一介绍

二多表连接查询

三符合条件连接查询

四子查询