day07-多表查询
本节重点:
- 多表连接查询
- 符合条件连接查询
- 子查询
准备工作:准备两张表,部门表(department)、员工表(employee)
create table department(
id int,
name varchar()
); create table employee(
id int primary key auto_increment,
name varchar(),
sex enum('male','female') not null default 'male',
age int,
dep_id int
); #插入数据
insert into department values
(,'技术'),
(,'人力资源'),
(,'销售'),
(,'运营'); insert into employee(name,sex,age,dep_id) values
('tom','male',,),
('mike','female',,),
('jack','male',,),
('lucy','female',,),
('lili','male',,),
('alice','female',,)
; # 查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int() | YES | | NULL | |
| name | varchar() | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+
rows in set (0.19 sec) mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int() | NO | PRI | NULL | auto_increment |
| name | varchar() | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int() | YES | | NULL | |
| dep_id | int() | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+
rows in set (0.01 sec) mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| | 技术 |
| | 人力资源 |
| | 销售 |
| | 运营 |
+------+--------------+
rows in set (0.02 sec) mysql> select * from employee;
+----+-------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+-------+--------+------+--------+
| | tom | male | | |
| | mike | female | | |
| | jack | male | | |
| | lucy | female | | |
| | lili | male | | |
| | alice | female | | |
+----+-------+--------+------+--------+
rows in set (0.00 sec) ps:观察两张表,发现department表中id=203部门在employee中没有对应的员工,发现employee中id=6的员工在department表中没有对应关系。
一、多表连接查询
两张表的准备工作已完成,比如现在我要查询的员工信息以及该员工所在的部门。从该题中,我们看出既要查员工又要查该员工的部门,肯定要将两张表进行连接查询。
重点:外链接语法
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;
1.1、交叉连接
不适用任何匹配条件。生成笛卡尔积(简单的说就是两个集合相乘的结果)。
mysql> select * from employee, department;
+----+-------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+-------+--------+------+--------+------+--------------+
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 1 | tom | male | 18 | 200 | 201 | 人力资源 |
| 1 | tom | male | 18 | 200 | 202 | 销售 |
| 1 | tom | male | 18 | 200 | 203 | 运营 |
| 2 | mike | female | 48 | 201 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 2 | mike | female | 48 | 201 | 202 | 销售 |
| 2 | mike | female | 48 | 201 | 203 | 运营 |
| 3 | jack | male | 38 | 201 | 200 | 技术 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 202 | 销售 |
| 3 | jack | male | 38 | 201 | 203 | 运营 |
| 4 | lucy | female | 28 | 202 | 200 | 技术 |
| 4 | lucy | female | 28 | 202 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 4 | lucy | female | 28 | 202 | 203 | 运营 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| 5 | lili | male | 18 | 200 | 201 | 人力资源 |
| 5 | lili | male | 18 | 200 | 202 | 销售 |
| 5 | lili | male | 18 | 200 | 203 | 运营 |
| 6 | alice | female | 18 | 204 | 200 | 技术 |
| 6 | alice | female | 18 | 204 | 201 | 人力资源 |
| 6 | alice | female | 18 | 204 | 202 | 销售 |
| 6 | alice | female | 18 | 204 | 203 | 运营 |
+----+-------+--------+------+--------+------+--------------+
24 rows in set (0.06 sec)
1.2、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;
或
mysql> select * from employee inner join department on employee.dep_id = department.id;
+----+------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+------+--------+------+--------+------+--------------+
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
+----+------+--------+------+--------+------+--------------+
5 rows in set (0.05 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;
1.3、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 * from employee left join department on employee.dep_id = department.id;
+----+-------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+----+-------+--------+------+--------+------+--------------+
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 6 | alice | female | 18 | 204 | NULL | NULL |
+----+-------+--------+------+--------+------+--------------+
6 rows in set (0.00 sec)
1.4、外链接之右连接:优先显示右表全部记录
mysql> select * from employee right join department on employee.dep_id = department.id;
+------+------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+------+------+--------+------+--------+------+--------------+
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| NULL | NULL | NULL | NULL | NULL | 203 | 运营 |
+------+------+--------+------+--------+------+--------------+
6 rows in set (0.00 sec)
1.5、全外连接:显示左右两个表全部记录(了解)
#外连接:在内连接的基础上增加左边和右边都有的结果
#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;
+------+-------+--------+------+--------+------+--------------+
| id | name | sex | age | dep_id | id | name |
+------+-------+--------+------+--------+------+--------------+
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 6 | alice | female | 18 | 204 | NULL | NULL |
| 1 | tom | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| NULL | NULL | NULL | NULL | NULL | 203 | 运营 |
+------+-------+--------+------+--------+------+--------------+
12 rows in set (0.00 sec) mysql> 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 | tom | male | 18 | 200 | 200 | 技术 |
| 5 | lili | male | 18 | 200 | 200 | 技术 |
| 2 | mike | female | 48 | 201 | 201 | 人力资源 |
| 3 | jack | male | 38 | 201 | 201 | 人力资源 |
| 4 | lucy | female | 28 | 202 | 202 | 销售 |
| 6 | alice | female | 18 | 204 | NULL | NULL |
| NULL | NULL | NULL | NULL | NULL | 203 | 运营 |
+------+-------+--------+------+--------+------+--------------+
7 rows in set (0.06 sec) #注意 union与union all的区别:union会去掉相同的纪录
二、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;
三、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
3.1、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);
+------+--------------+
| id | name |
+------+--------------+
| 201 | 人力资源 |
| 202 | 销售 |
+------+--------------+ # 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');
+------+
| name |
+------+
| tom |
| lili |
+------+ #查看不足1人的部门名
select name from department
where id not in
(select dep_id from employee group by dep_id); 或
mysql> select name from department
where id not in
(select dep_id from employee);
+--------+
| name |
+--------+
| 运营 |
+--------+
1 row in set (0.00 sec)
3.2、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 employee where age > (select avg(age) from employee);
+------+------+
| name | age |
+------+------+
| mike | 48 |
| jack | 38 |
+------+------+ #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,再对根据临时表的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。 mysql> select name,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 |
+------+------+
| mike | 48 |
+------+------+
1 row in set (0.07 sec)
3.3、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=200,True
mysql> select * from employee where exists (select id from department where id=200);
+----+-------+--------+------+--------+
| id | name | sex | age | dep_id |
+----+-------+--------+------+--------+
| 1 | tom | male | 18 | 200 |
| 2 | mike | female | 48 | 201 |
| 3 | jack | male | 38 | 201 |
| 4 | lucy | female | 28 | 202 |
| 5 | lili | male | 18 | 200 |
| 6 | alice | female | 18 | 204 |
+----+-------+--------+------+--------+ #department表中不存在dept_id=204,False
mysql> select * from employee where exists (select id from department where id=204);
Empty set (0.00 sec)
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