主要内容
- 1.表和表的关系
- 2.单表查询
- 3.多表查询
1.表和表的关系(外键的变种)
分析步骤： #1、先站在左表的角度去找 是否左表的多条记录可以对应右表的一条记录，如果是，则证明左表的一个字段foreign key 右表一个字段（通常是id） #2、再站在右表的角度去找 是否右表的多条记录可以对应左表的一条记录，如果是，则证明右表的一个字段foreign key 左表一个字段（通常是id） #3、总结： #多对一： 如果只有步骤1成立，则是左表多对一右表 如果只有步骤2成立，则是右表多对一左表 #多对多 如果步骤1和2同时成立，则证明这两张表时一个双向的多对一，即多对多,需要定义一个这两张表的关系表来专门存放二者的关系 #一对一: 如果1和2都不成立，而是左表的一条记录唯一对应右表的一条记录，反之亦然。这种情况很简单，就是在左表foreign key右表的基础上，将左表的外键字段设置成unique即可
3.多表查询
准备两张表: 部门表（department）、员工表（employee）
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), ('nvshen','male',18,200), ('xiaomage','female',18,204) ; # 查看表结构和数据 mysql> desc department; +-------+-------------+------+-----+---------+-------+ | Field | Type | Null | Key | Default | Extra | +-------+-------------+------+-----+---------+-------+ | id | int(11) | YES | | NULL | | | name | varchar(20) | YES | | NULL | | +-------+-------------+------+-----+---------+-------+ rows in set (0.19 sec) mysql> desc employee; +--------+-----------------------+------+-----+---------+----------------+ | Field | Type | Null | Key | Default | Extra | +--------+-----------------------+------+-----+---------+----------------+ | id | int(11) | NO | PRI | NULL | auto_increment | | name | varchar(20) | YES | | NULL | | | sex | enum('male','female') | NO | | male | | | age | int(11) | YES | | NULL | | | dep_id | int(11) | YES | | NULL | | +--------+-----------------------+------+-----+---------+----------------+ rows in set (0.01 sec) mysql> select * from department; +------+--------------+ | id | name | +------+--------------+ | 200 | 技术 | | 201 | 人力资源 | | 202 | 销售 | | 203 | 运营 | +------+--------------+ rows in set (0.02 sec) mysql> select * from employee; +----+----------+--------+------+--------+ | 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 | +----+----------+--------+------+--------+ rows in set (0.00 sec)
(1)多表连接查询
笛卡尔积:
mysql> select * from employee2,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 | nvshen | male | 18 | 200 | 200 | 技术 | | 5 | nvshen | male | 18 | 200 | 201 | 人力资源 | | 5 | nvshen | male | 18 | 200 | 202 | 销售 | | 5 | nvshen | male | 18 | 200 | 203 | 运营 | | 6 | xiaomage | female | 18 | 204 | 200 | 技术 | | 6 | xiaomage | female | 18 | 204 | 201 | 人力资源 | | 6 | xiaomage | female | 18 | 204 | 202 | 销售 | | 6 | xiaomage | female | 18 | 204 | 203 | 运营 | +----+----------+--------+------+--------+------+--------------+ 24 rows in set (0.11 sec)
#符合条件查询 mysql> select * from employee2,department where employee2.dep_id = department.id; +----+---------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+---------+--------+------+--------+------+--------------+ | 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 | nvshen | male | 18 | 200 | 200 | 技术 | +----+---------+--------+------+--------+------+--------------+
(2) 内连接 (只获取匹配数据)
#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 * from employee2 inner join department on employee2.dep_id = department.id; +----+---------+--------+------+--------+------+--------------+ | id | name | sex | age | dep_id | id | name | +----+---------+--------+------+--------+------+--------------+ | 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 | nvshen | male | 18 | 200 | 200 | 技术 | +----+---------+--------+------+--------+------+--------------+
(3)左连接:优先显示左表所有记录
mysql> select *from employee2 left join department on employee2.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 | +----+----------+--------+------+--------+------+--------------+
(4)全外连接
mysql> select * from employee2 left join department on employee2.dep_id = department.id -> union -> select * from employee2 right join department on employee2.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)
练习题:
#1.即找出年龄大于25岁的员工以及员工所在的部门 mysql> select department.name,employee2.name from employee2 inner join department on employee2.dep_id=department.id where age>25; +--------------+---------+ | name | name | +--------------+---------+ | 人力资源 | alex | | 人力资源 | wupeiqi | | 销售 | yuanhao | +--------------+---------+
(5)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
小练习:
# 查询大于部门内平均年龄的员工名、年龄 select name,age from employee2 inner join (select dep_id,avg(age) as av from employee2
group by dep_id) as B on employee2.dep_id = B.dep_id where age > av;
#查询平均年龄在25岁以上的部门名 select * from department where id in (select dep_id from employee group by dep_id having avg(age) > 25);