时间:2021-07-01 10:21:17 帮助过:8人阅读
- # 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;
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
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;
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)
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)
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
- # 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)
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)
【MySQL】多表查询
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