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  • mysql之多表查询

    多表查询 CSDN

    目录

    多表链接查询

    笛卡尔积

    内链接 inner join

    外链接之左链接 left join

    外链接之右链接 right join

    全外链接

    符合条件链接查询

    子查询


    先准备两张表:部门表(department)、员工表(employee)

    # 部门表
    create table department(
        id int primary key auto_increment,
        name varchar(20) not null
        );
    
    # 员工表
    create table employee(
        id int primary key auto_increment,
        name varchar(20) not null,
        sex enum('male', 'female') not null default 'male',
        age int not null,
        dep_id int not null
        );
    
    # 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),
    ('nvshen', 'male', 18, 200),
    ('xiaomage', 'female', 18, 204)
    ;
    
    # 注意:
    department表中id=203的部门在employee中没有对应的员工
    employee表中id=6的员工在department表中没有对应的部门

    多表链接查询

    # 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;

    笛卡尔积:

    即交叉链接
    不适用任何匹配条件,生成笛卡尔积关于笛卡尔积的含义,请自行百度)

    select * from employee, department;    # 语法

    内链接 inner join

    只链接匹配的行

    # 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匹配的结果
    # 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 | nvshen  |  18 | male   | 技术         |
    +----+---------+-----+--------+--------------+
    5 rows in set (0.00 sec)
    
    # 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;

    外链接之左链接 left join

    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 | nvshen   | 技术         |
    |  2 | alex     | 人力资源     |
    |  3 | wupeiqi  | 人力资源     |
    |  4 | yuanhao  | 销售         |
    |  6 | xiaomage | NULL         |
    +----+----------+--------------+
    6 rows in set (0.00 sec)

    外链接之右链接 right join

    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 | nvshen  | 技术         |
    | NULL | NULL    | 运营         |
    +------+---------+--------------+
    6 rows in set (0.00 sec)

    全外链接

    显示左右两个表全部记录

    # 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
    
    mysql> select * from employee left join department 
        -> on employee.dep_id = department.id 
        -> union all
        -> select * from employee right join department 
        -> on employee.dep_id = department.id;
    
    mysql> 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;
    +------+----------+--------+------+--------+------+--------------+
    | id   | name     | sex    | age  | dep_id | id   | name         |
    +------+----------+--------+------+--------+------+--------------+
    |    1 | egon     | male   |   18 |    200 |  200 | 技术         |
    |    5 | nvshen   | 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 | xiaomage | female |   18 |    204 | NULL | NULL         |
    | NULL | NULL     | NULL   | NULL |   NULL |  203 | 运营         |
    +------+----------+--------+------+--------+------+--------------+
    7 rows in set (0.00 sec)
    
    # 注意 union与union all的区别:union会去掉相同的纪录

    符合条件链接查询

    # 示例1 以内链接的方式查询:找出年龄大于25岁的员工以及员工所在的部门
    mysql> select employee.name,employee.age,department.name
        -> from employee inner join department
        -> on employee.dep_id = department.id
        -> where age > 25;
    +---------+-----+--------------+
    | name    | age | name         |
    +---------+-----+--------------+
    | alex    |  48 | 人力资源     |
    | wupeiqi |  38 | 人力资源     |
    | yuanhao |  28 | 销售         |
    +---------+-----+--------------+
    3 rows in set (0.00 sec)
    
    
    # 示例2 以内链接的方式查询:以age字段的升序方式显示
    mysql> select employee.name,employee.age,department.name
        -> from employee inner join department
        -> on employee.dep_id = department.id
        -> order by age asc;    # 升序排序
    +---------+-----+--------------+
    | name    | age | name         |
    +---------+-----+--------------+
    | egon    |  18 | 技术         |
    | nvshen  |  18 | 技术         |
    | yuanhao |  28 | 销售         |
    | wupeiqi |  38 | 人力资源     |
    | alex    |  48 | 人力资源     |
    +---------+-----+--------------+
    5 rows in set (0.00 sec)

    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、not exists 等关键字.
    4. 还可以包含比较运算符:=、 !=、>、< 等.

    示例1:带in关键字的子查询

    # 查询平均年龄在25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 id,name from employee
        where dep_id in
            (select id from department where name="技术");
    
    # 查无人的部门名
    select name from department
        where id not in
            (select dep_id from employee);

    示例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
    select name,age from employee 
        where age > (select avg(age) from employee);
    
    # 查询大于部门内平均年龄的员工名、年龄
    思路:
        (1)先对员工表(employee)中的人员分组(group by),查询出dep_id以及平均年龄。
         (2)将查出的结果作为临时表,再根据临时表的dep_id和employee的dep_id作为筛选条件将employee表和临时表进行内连接。
         (3)最后再将employee员工的年龄是大于平均年龄的员工名字和年龄筛选。
    
    select t1.name,t1.age from employee as t1
        inner join
        (select dep_id,avg(age) as avg_age from employee group by dep_id) as t2
        on t1.dep_id = t2.dep_id
        where t1.age > t2.avg_age;
    
    +------+-----+
    | name | age |
    +------+-----+
    | alex |  48 |
    +------+-----+
    1 row in set (0.00 sec)

    示例3:带exists关键字的子查询

    # exists关键字表示存在。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
    
    # exists为True时:
    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 | nvshen   | male   |  18 |    200 |
    |  6 | xiaomage | female |  18 |    204 |
    +----+----------+--------+-----+--------+
    6 rows in set (0.00 sec)
    
    # exists为False时:
    mysql> select * from employee where exists (select id from department where id=204);
    Empty set (0.00 sec)

     

     



     

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  • 原文地址:https://www.cnblogs.com/babyjoy/p/9655358.html
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