• day 40 多表查询 子查询


    一、本节介绍:

    本节主题

    • 多表连接查询
    • 复合条件连接查询
    • 子查询

    准备表

    #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 |
    +----+------------+--------+------+--------+
    插入的两个表

    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;

    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;
    View Code

    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
    View Code

    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
    select name from department
        where id in 
            (select dep_id from employee group by dep_id having count(id) <=1);

    2 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; 

    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

    #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)

    练习:查询每个部门最新入职的那位员工

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  • 原文地址:https://www.cnblogs.com/number1994/p/8290096.html
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