多表查询

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)
;
练习表

一、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:不使用任何匹配条件,生成笛卡尔积

 积 表示乘积的意思
    把两个表中的所有数据 全部建立关联关系
    a 表 有二条  b表有三条   总数据量为2 * 3 = 6条

    可以保证 肯定有一条关联关系是正确的,但是同时会产生大量错误数据,
    我们需要加以过滤 来得到正确的数据

------------------------------------------
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         |
+----+------------+--------+------+--------+------+--------------+
View Code

2、内连接:只连接匹配的

两边数据完全匹配成功才会显示
inner(可以省略) join  on(on只能与join一起使用,但是在join中 可以把on换成where)
on == where    只能用于连接查询
如果用来筛选匹配关系  建议使用on  连接查询中必须使用on
mysql> select *from department,employee where dep_id = department.id and department.name = "技术";
+------+--------+----+-----------+------+------+--------+
| id   | name   | id | name      | sex  | age  | dep_id |
+------+--------+----+-----------+------+------+--------+
|  200 | 技术   |  1 | egon      | male |   18 |    200 |
|  200 | 技术   |  5 | liwenzhou | male |   18 |    200 |
+------+--------+----+-----------+------+------+--------+

#这种方式与上面的方式类似,相当于,join连接两个表,这里的on相当于上边的where
mysql> select *from department join employee on dep_id = department.id where department.name = "技术";
+------+--------+----+-----------+------+------+--------+
| id   | name   | id | name      | sex  | age  | dep_id |
+------+--------+----+-----------+------+------+--------+
|  200 | 技术   |  1 | egon      | male |   18 |    200 |
|  200 | 技术   |  5 | liwenzhou | male |   18 |    200 |
+------+--------+----+-----------+------+------+--------+
View Code
#多对多:
    1.先把三个表全都连在一起 select * from stu join tsr join tea
    2.用on来筛选出正确关系 on stu.id = tsr.s_id and tea.id = tsr.t_id         (把错误的丢掉)
    3.然后通过where 添加额外的条件 where tea.name = "egon";

stu tsr join 为三个表

3、外连接之左右连接


#外连接查询结果=内连接查询结果+主表中有而从表中没有的
#外链接之左连接:(左边的是主表) 左边无论是否匹配,都全部显示,右边只显示匹配成功的 #语法: select * from 表1 left join 表2 on 条件 select * from employee left join department on dep_id = department.id;

4、全外连接

#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
;

##注意 union与union all的区别:union会去掉相同的纪录
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 | 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         |
+------+------------+--------+------+--------+------+--------------+
7 rows in set (1.58 sec)

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;
+------+------------+--------+------+--------+------+--------------+
| 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         |
|    1 | egon       | 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         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+
12 rows in set (0.00 sec)
View Code

5、非等值连接查询

#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;
View Code

 ----------------------------------------

1、内链接

2、左右连接

3、左右连接升级版

 

4、全连接,MySQL不支持full,  使用左右连接拼接

 

5、使用左右连接升级版拼接

 

 

二、子查询

create table emp (id int,name char(10),sex char,age int,dept_id int,job char(10),salary double);

insert into emp values

(1,"刘备","",26,1,"总监",5800),

(2,"张飞","",24,1,"员工",3000),

(3,"关羽","",30,1,"员工",4000),

(4,"孙权","",25,2,"总监",6000),

(5,"周瑜","",22,2,"员工",5000),

(6,"小乔","",31,2,"员工",4000),

(7,"曹操","",19,3,"总监",10000),

(8,"司马懿","",24,3,"员工",6000);



create table dept(id int primary key,name char(10));

insert into dept values(1,"市场"),(2,"行政"),(3,"财务");
练习表
#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

过程推导

#练习,查查财务部有哪些人,

#连接查询:
mysql> select *from dept join emp
    -> on dept_id = dept.id
    -> where dept.name = "财务";

#子查询:
先通过部门名称拿到部门id 
select name from emp where dept_id = 3;
在根据id 找对应的员工
select name from emp where dept_id = (select id from dept where name = "财务");


#连到一起
select name from emp where dept_id = (select id from dept where name = "财务");

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);
View Code

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; 
View Code

3、带EXISTS关键字的子查询

#exists(相关子查询)

#查询到返回1,查询不到返回0

select exists(select *id from emp where name='张三')
原文地址:https://www.cnblogs.com/pdun/p/11333605.html