标签:nio order by 员工 dep 现在 计算 连表 lex 筛选
# 笛卡尔积
# 先计算两张表的笛卡尔积,再根据用户给出的条件进行筛选
select * from employee,department where dep_id = department.id
# 内连接 inner join ... on 连接条件
select * from 表1 inner join 表2 on 条件
select * from employee inner join department on dep_id = department.id
employee --> dep_id: 200,201,202,204
department --> id : 200,201,202,203
# 内连接 :只显示两张表中互相匹配的项,其他不匹配的不显示
# 外连接z
# 左外连接 left join .. on
select * from 表1 left join 表2 on 条件
select * from employee left join department on dep_id = department.id
select * from department left join employee on dep_id = department.id
# 不管左表中是不是匹配上都会显示所有内容
# 右外连接 right join .. on
select * from 表1 right join 表2 on 条件
select * from employee right join department on dep_id = department.id
# 不管右表中是不是匹配上都会显示所有内容
# 全外连接
select * from department left join employee on dep_id = department.id
union
select * from department right join employee on dep_id = department.id
# 连表查询
# 原来是两张表,现在拼成一张表,所有的操作都像是操作一张表一样了
# 年龄大于25岁的员工以及员工所在的部门
select * from employee inner join department d on d.id = dep_id;
# 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 * from employee e inner join department d on d.id = e.dep_id order by age;
# 存在年龄大于25岁员工的部门
select * from employee where age>25;
select distinct dep_id from employee where age>25;
select * from department where id in (select distinct dep_id from employee where age>25);
# a.查询平均年龄在25岁以上的部门名
# 求部门的平均年龄
select avg(age) from employee group by dep_id
# 求部门的平均年龄>25岁的 部门id
select dep_id from employee group by dep_id having avg(age)>25
# 根据部门id求部门名
select name from department where id in (select dep_id from employee group by dep_id having avg(age)>25);
# b.查询平均年龄在25岁以上的部门名,平均年龄的值
# 方法1:
# 先查部门的平均年龄>25岁的部门id,平均年龄
select dep_id,avg(age) from employee group by dep_id having avg(age) > 25
# 查出结果在之后再连表
select name,avg_age from department as d right join (select dep_id,avg(age) as avg_age from employee group by dep_id having avg(age) > 25) as t on d.id = t.dep_id
# 方法2
# 先查各部门的平均年龄
select dep_id,avg(age) from employee group by dep_id;
# 查出结果在之后再连表
select name,avg_age from
department as d inner join (select dep_id,avg(age) as avg_age from employee group by dep_id) as t
on d.id =t.dep_id
where avg_age>25
# 根据a,b总结出来的规律:
# 如果最终需要的结果只出现在一张表中,可以用子查询解决问题
# 如果最终需要的结果出现在2表中,那么最后用的一定是连表查询
# 查看"技术"部员工姓名
# 技术部门在department表中的id
select id from department where name = '技术';
# 查的是姓名 employee
select name from employee where dep_id = (select id from department where name = '技术');
# 查看不足1人的部门名
# 子查询
# 先查出所有人的部门id
select distinct dep_id from employee;
# 看不在部门id这个集合中的部门有哪些
select name from department where id not in (select distinct dep_id from employee);
# 连表
select d.name from department d left join employee e on d.id = dep_id where e.id is null;
# 查询大于所有人平均年龄的员工名与年龄
# 先查询所有人的平均年龄
select avg(age) from employee;
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
# 先求各部门的平均年龄
select dep_id,avg(age) from employee group by dep_id;
# 拼接
select e.name,e.age from
employee e
inner join
(select dep_id,avg(age) avg_age from employee group by dep_id) t
on e.dep_id = t.dep_id
where e.age > t.avg_age;
# dep_id
# 201
# 202
# 204
in ;not in
# id name
# 1 alex
select name from emp;
select name as n,(select age from employee where name = n) from emp;
标签:nio order by 员工 dep 现在 计算 连表 lex 筛选
原文地址:https://www.cnblogs.com/lyoko1996/p/11328913.html