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  • 数据库——数据操作——多表查询（10）

    • 1.准备两个表
    •  1 #建表
 2 create table department(
 3 id int,
 4 name varchar(20) 
 5 );
 6 
 7 create table employee(
 8 id int primary key auto_increment,
 9 name varchar(20),
10 sex enum('male','female') not null default 'male',
11 age int,
12 dep_id int
13 );
14 
15 #插入数据
16 insert into department values
17 (200,'技术'),
18 (201,'人力资源'),
19 (202,'销售'),
20 (203,'运营');
21 
22 insert into employee(name,sex,age,dep_id) values
23 ('egon','male',18,200),
24 ('alex','female',48,201),
25 ('wupeiqi','male',38,201),
26 ('yuanhao','female',28,202),
27 ('liwenzhou','male',18,200),
28 ('jingliyang','female',18,204)
29 ;
30 
31 
32 #查看表结构和数据
33 mysql> desc department;
34 +-------+-------------+------+-----+---------+-------+
35 | Field | Type | Null | Key | Default | Extra |
36 +-------+-------------+------+-----+---------+-------+
37 | id | int(11) | YES | | NULL | |
38 | name | varchar(20) | YES | | NULL | |
39 +-------+-------------+------+-----+---------+-------+
40 
41 mysql> desc employee;
42 +--------+-----------------------+------+-----+---------+----------------+
43 | Field | Type | Null | Key | Default | Extra |
44 +--------+-----------------------+------+-----+---------+----------------+
45 | id | int(11) | NO | PRI | NULL | auto_increment |
46 | name | varchar(20) | YES | | NULL | |
47 | sex | enum('male','female') | NO | | male | |
48 | age | int(11) | YES | | NULL | |
49 | dep_id | int(11) | YES | | NULL | |
50 +--------+-----------------------+------+-----+---------+----------------+
51 
52 mysql> select * from department;
53 +------+--------------+
54 | id | name |
55 +------+--------------+
56 | 200 | 技术 |
57 | 201 | 人力资源 |
58 | 202 | 销售 |
59 | 203 | 运营 |
60 +------+--------------+
61 
62 mysql> select * from employee;
63 +----+------------+--------+------+--------+
64 | id | name | sex | age | dep_id |
65 +----+------------+--------+------+--------+
66 | 1 | egon | male | 18 | 200 |
67 | 2 | alex | female | 48 | 201 |
68 | 3 | wupeiqi | male | 38 | 201 |
69 | 4 | yuanhao | female | 28 | 202 |
70 | 5 | liwenzhou | male | 18 | 200 |
71 | 6 | jingliyang | female | 18 | 204 |
72 +----+------------+--------+------+--------+
73 
74 表department与employee
      准备表
    • 2.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
      •  1 mysql> select * from employee,department;
 2 +----+------------+--------+------+--------+------+--------------+
 3 | id | name       | sex    | age  | dep_id | id   | name         |
 4 +----+------------+--------+------+--------+------+--------------+
 5 |  1 | egon       | male   |   18 |    200 |  200 | 技术         |
 6 |  1 | egon       | male   |   18 |    200 |  201 | 人力资源     |
 7 |  1 | egon       | male   |   18 |    200 |  202 | 销售         |
 8 |  1 | egon       | male   |   18 |    200 |  203 | 运营         |
 9 |  2 | alex       | female |   48 |    201 |  200 | 技术         |
10 |  2 | alex       | female |   48 |    201 |  201 | 人力资源     |
11 |  2 | alex       | female |   48 |    201 |  202 | 销售         |
12 |  2 | alex       | female |   48 |    201 |  203 | 运营         |
13 |  3 | wupeiqi    | male   |   38 |    201 |  200 | 技术         |
14 |  3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
15 |  3 | wupeiqi    | male   |   38 |    201 |  202 | 销售         |
16 |  3 | wupeiqi    | male   |   38 |    201 |  203 | 运营         |
17 |  4 | yuanhao    | female |   28 |    202 |  200 | 技术         |
18 |  4 | yuanhao    | female |   28 |    202 |  201 | 人力资源     |
19 |  4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
20 |  4 | yuanhao    | female |   28 |    202 |  203 | 运营         |
21 |  5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
22 |  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
23 |  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
24 |  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
25 |  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
26 |  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
27 |  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
28 |  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
29 +----+------------+--------+------+--------+------+--------------+
        View Code
      • 2.内连接：只连接匹配的行（inner join）
      •  1 #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
 2 #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
 3 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
 4 +----+-----------+------+--------+--------------+
 5 | id | name      | age  | sex    | name         |
 6 +----+-----------+------+--------+--------------+
 7 |  1 | egon      |   18 | male   | 技术         |
 8 |  2 | alex      |   48 | female | 人力资源     |
 9 |  3 | wupeiqi   |   38 | male   | 人力资源     |
10 |  4 | yuanhao   |   28 | female | 销售         |
11 |  5 | liwenzhou |   18 | male   | 技术         |
12 +----+-----------+------+--------+--------------+
13 
14 #上述sql等同于
15 mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
        View Code
      • 3.外连接之左连接：优先显示左表全部记录（left join），在内连接的基础上保留左表的记录
      •  1 #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
 2 #本质就是：在内连接的基础上增加左边有右边没有的结果
 3 mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
 4 +----+------------+--------------+
 5 | id | name       | depart_name  |
 6 +----+------------+--------------+
 7 |  1 | egon       | 技术         |
 8 |  5 | liwenzhou  | 技术         |
 9 |  2 | alex       | 人力资源     |
10 |  3 | wupeiqi    | 人力资源     |
11 |  4 | yuanhao    | 销售         |
12 |  6 | jingliyang | NULL         |
13 +----+------------+--------------+
        View Code
      • 4.外连接之右连接：优先显示右表全部记录（right join），在内连接的基础上保留右表的记录
      •  1 #以右表为准，即找出所有部门信息，包括没有员工的部门
 2 #本质就是：在内连接的基础上增加右边有左边没有的结果
 3 mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
 4 +------+-----------+--------------+
 5 | id   | name      | depart_name  |
 6 +------+-----------+--------------+
 7 |    1 | egon      | 技术         |
 8 |    2 | alex      | 人力资源     |
 9 |    3 | wupeiqi   | 人力资源     |
10 |    4 | yuanhao   | 销售         |
11 |    5 | liwenzhou | 技术         |
12 | NULL | NULL      | 运营         |
13 +------+-----------+--------------+
        View Code
      • 5.全外连接：显示左右两个表全部记录（left join。。。union。。。right join。。。），在内连接的基础上保留左右两表没有对应关系的记录
      •  1 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
 2 #注意：mysql不支持全外连接 full JOIN
 3 #强调：mysql可以使用此种方式间接实现全外连接
 4 select * from employee left join department on employee.dep_id = department.id
 5 union
 6 select * from employee right join department on employee.dep_id = department.id
 7 ;
 8 #查看结果
 9 +------+------------+--------+------+--------+------+--------------+
10 | id   | name       | sex    | age  | dep_id | id   | name         |
11 +------+------------+--------+------+--------+------+--------------+
12 |    1 | egon       | male   |   18 |    200 |  200 | 技术         |
13 |    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
14 |    2 | alex       | female |   48 |    201 |  201 | 人力资源     |
15 |    3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
16 |    4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
17 |    6 | jingliyang | female |   18 |    204 | NULL | NULL         |
18 | NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
19 +------+------------+--------+------+--------+------+--------------+
20 
21 #注意 union与union all的区别：union会去掉相同的纪录
        View Code
    • 3.符合条件连接查询 
      •  1 #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
 2 select employee.name,department.name from employee inner join department
 3     on employee.dep_id = department.id
 4     where age > 25;
 5 
 6 #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
 7 select employee.id,employee.name,employee.age,department.name from employee,department
 8     where employee.dep_id = department.id
 9     and age > 25
10     order by age asc;
        View Code
    • 4.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
      •  1 #查询平均年龄在25岁以上的部门名
 2 select id,name from department
 3     where id in 
 4         (select dep_id from employee group by dep_id having avg(age) > 25);
 5 
 6 #查看技术部员工姓名
 7 select name from employee
 8     where dep_id in 
 9         (select id from department where name='技术');
10 
11 #查看不足1人的部门名
12 select name from department
13     where id in 
14         (select dep_id from employee group by dep_id having count(id) <=1);
        View Code
      • 2.带比较运算符的子查询
      •  1 #比较运算符：=、!=、>、>=、<、<=、<>
 2 #查询大于所有人平均年龄的员工名与年龄
 3 mysql> select name,age from emp where age > (select avg(age) from emp);
 4 +---------+------+
 5 | name | age |
 6 +---------+------+
 7 | alex | 48 |
 8 | wupeiqi | 38 |
 9 +---------+------+
10 rows in set (0.00 sec)
11 
12 
13 #查询大于部门内平均年龄的员工名、年龄
14 select t1.name,t1.age from emp t1
15 inner join 
16 (select dep_id,avg(age) avg_age from emp group by dep_id) t2
17 on t1.dep_id = t2.dep_id
18 where t1.age > t2.avg_age;
        View Code
      • 3.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
         1 #department表中存在dept_id=203，Ture
 2 mysql> select * from employee
 3     ->     where exists
 4     ->         (select id from department where id=200);
 5 +----+------------+--------+------+--------+
 6 | id | name       | sex    | age  | dep_id |
 7 +----+------------+--------+------+--------+
 8 |  1 | egon       | male   |   18 |    200 |
 9 |  2 | alex       | female |   48 |    201 |
10 |  3 | wupeiqi    | male   |   38 |    201 |
11 |  4 | yuanhao    | female |   28 |    202 |
12 |  5 | liwenzhou  | male   |   18 |    200 |
13 |  6 | jingliyang | female |   18 |    204 |
14 +----+------------+--------+------+--------+
15 
16 #department表中存在dept_id=205，False
17 mysql> select * from employee
18     ->     where exists
19     ->         (select id from department where id=204);
20 Empty set (0.00 sec)
        View Code
    • 5.综合练习
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  • 原文地址：https://www.cnblogs.com/GraceZ/p/8452901.html
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