# 一 \ 内连接
# select * from department
# inner join employee
# on department.id = employee.dep_id;
# 只会把左表和右表对应的行显示出来
# 如果左表中的department.id在employee.dep_id中没有出现,那么会抛弃这个行
# 如果右表中的employee.dep_id在department.id中没有出现,那么也会抛弃这个行# 内连接的另一种方式 用where
# select * from department,employee
# where department.id = employee.dep_id;# 从不同的表中取字段
# select employee.name,employee.sex,department.name
# from department,employee
# where department.id = employee.dep_id;# 给表起别名,让sql更简单
# select emp.name,emp.sex,dep.name
# from department as dep,employee as emp
# where dep.id = emp.dep_id;# 给字段起别名,让显示的效果更明确
# select emp.name,emp.sex,dep.name as dep_name
# from department as dep,employee as emp
# where dep.id = emp.dep_id;# 外连接
# 二 \左外连接 ： 显示左表中的所有项，和右表中所有满足拼接条件的项
# select * from department
# left join employee
# on department.id = employee.dep_id;# 三 \右外连接：显示右表中的所有项，和左表中所有满足拼接条件的项
# select * from department
# right join employee
# on department.id = employee.dep_id;# 四 \全外连接：左表和右表都完全显示出来了
# select * from department left join employee on department.id = employee.dep_id
# union all
# select * from department right join employee on department.id = employee.dep_id# 练习
# 示例一 : 找到年龄> 25的员工的姓名和部门
# select * from employee
# inner join department
# on employee.dep_id = department.id
# where age>25;# 示例一变式 : 找到alex员工的年龄和部门
# select age,department.name from employee
# inner join department
# on employee.dep_id = department.id
# where employee.name = 'alex';# 给表重命名
# select age,dep.name from employee as emp
# inner join department as dep
# on emp.dep_id = dep.id
# where emp.name = 'alex';# 给字段重命名引起的错误
# select employee.name as emp_name,age,department.name from employee
# inner join department
# on employee.dep_id = department.id
# where emp_name = 'alex';
# 报错,因为在select处重名名不能在where/group by/having中使用,由于mysql的词法分析顺序导致该问题# 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
# select * from employee
# inner join department
# on employee.dep_id = department.id
# order by age;

# 示例一 : 查询平均年龄在25岁以上的部门名
# 涉及到年龄 员工表
# 部门名字   部门表
# 你的结果在哪个表,那个表一定不是子查询的表# 1. 连表查询的结果
# 先内连接得到一张大表
# select * from department
# inner join employee
# on department.id = employee.dep_id# 再根据部门分组
# select department.name from department
# inner join employee
# on department.id = employee.dep_id
# group by department.id
# having avg(age) > 25;# 2. 子查询的结果
# 先完成一部分需求,求每一个部门的人的平均年龄
# select dep_id,avg(age) from employee group by dep_id
# 再筛选出平均年龄大于25的部门
# select dep_id,avg(age) from employee group by dep_id having avg(age)>25
# 由于我们只需要部门名称,而和部门名称相关的项就只有部门id,所以我们只留下dep_id字段
# select dep_id from employee group by dep_id having avg(age)>25
# 查询部门表,找到id在上面这个查询结果内的内容
# select name from department where id in (
# select dep_id from employee group by dep_id having avg(age)>25
# )
pass
# 示例2 : 查看"技术"部员工姓名
# 结果是 : 姓名 -  员工表
# 怎么知道技术部是谁? 怎么和员工表关联?
# 如果我能知道技术部的id是多少,就可以查询了
# 1.查询技术部的id
# select id from department where name = '技术'
# 2.取id=200的所有人
# select * from employee where dep_id = (
# select id from department where name = '技术');
# 3.只取名字
# select name from employee where dep_id = (
# select id from department where name = '技术');
pass
# 示例3 :查看不足1人的部门名
# 结果是 部门名 - 部门表
# 先操作员工表
# 1.找到员工表中所有人的部门id
# select dep_id from employee group by dep_id;
# select distinct dep_id from employee;
# 2.操作部门表查看id not in 上面范围中的项目
# select name from department where id not in (
# select dep_id from employee group by dep_id);# select name from department where id not in (
# select distinct dep_id from employee);