How to implement multi-table query in MySQL? MySQL multi-table query statements

The content of this article is to introduce how MySQL implements multi-table queries? MySQL multi-table query statements. It has certain reference value. Friends in need can refer to it. I hope it will be helpful to you.

Create table

# 创建表
create table department(id int,name varchar(20));
create table employee1(
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 employee1(name,sex,age,dep_id) values('egon','male',18,200),('alex','female',48,201),('tom','male',38,201),('yuanhao','female',28,202),('lidawei','male',18,200),('jinkezhou','female',18,204);

# 查看表
mysql> select * from employee1;
+----+-----------+--------+------+--------+
| id | name      | sex    | age  | dep_id |
+----+-----------+--------+------+--------+
|  1 | egon      | male   |   18 |    200 |
|  2 | alex      | female |   48 |    201 |
|  3 | tom       | male   |   38 |    201 |
|  4 | yuanhao   | female |   28 |    202 |
|  5 | lidawei   | male   |   18 |    200 |
|  6 | jinkezhou | female |   18 |    204 |
+----+-----------+--------+------+--------+
6 rows in set (0.00 sec)
mysql> select * from department;
+------+--------------+
| id   | name         |
+------+--------------+
|  200 | 技术       |
|  201 | 人力资源   |
|  202 | 销售       |
|  203 | 运营       |
+------+--------------+
4 rows in set (0.00 sec)

Multiple table connection query

Cross connection

Cross-join: No matching conditions apply. Generate Cartesian product

mysql> select * from employee1 ,department;

Inner join

Inner join: Find the common parts of the two tables, which is equivalent to using conditions to derive the Cartesian product The correct results were filtered out from the results. (Only connect matching rows)

# 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 * from employee1,department where employee1.dep_id=department.id;

#上面用where表示的可以用下面的内连接表示，建议使用下面的那种方法
mysql> select * from employee1 inner join department on employee1.dep_id=department.id;

# 也可以这样表示哈
mysql> select employee1.id,employee1.name,employee1.age,employee1.sex,department.name from employee1,department where employee1.dep_id=department.id;

Left join left

Prioritizes to display all records in the left table.

#左链接：在按照on的条件取到两张表共同部分的基础上，保留左表的记录
mysql> select * from employee1 left join department on department.id=employee1.dep_id;

mysql> select * from department left join  employee1 on department.id=employee1.dep_id;

Right joinright

Displays all records in the right table first.

#右链接：在按照on的条件取到两张表共同部分的基础上，保留右表的记录
mysql> select * from employee1 right join department on department.id=employee1.dep_id;
mysql> select * from department right join employee1 on department.id=employee1.dep_id;

All join join

mysql> select * from department full join employee1;

Multiple table query that meets the conditions

Example 1: Query the employee and department tables using inner joins, and the age field value in the employee table must be greater than 25,
That is, find employees who are older than 25 years old in all departments of the company

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and age>25;

Example 2: Query the employee and department tables using inner joins, and display the age field in ascending order

mysql> select * from employee1 inner join department on employee1.dep_id=department.id and age>25 and age>25 order by age asc;

Subquery

#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：= 、 !=、> 、<p> Example : </p><pre class="brush:php;toolbar:false"># 查询平均年龄在25岁以上的部门名
mysql> select name from department where id in ( select dep_id from employee1 group by dep_id having avg(age) > 25 );

# 查看技术部员工姓名
mysql> select name from employee1 where dep_id = (select id from department where name='技术');

# 查看小于2人的部门名
mysql> select name from department where id in (select dep_id from employee1 group by dep_id having count(id)  select * from department where id not in (select distinct dep_id from employee1);

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