LeetCode MySQL刷题——day1

目录

​​一、组合两个表​​

​​1、题目描述​​

​​2、题解​​

​​ 3、源码​​

​​二、超过经理收入的员工​​

​​1、题目描述​​

​​2、题解​​

​​3、源码​​

​​ 三、查找重复的电子邮箱​​

​​1、题目描述​​

​​2、题解​​

​​3、源码​​

​​ 四、从不订购的客户​​

​​1、题目描述​​

​​2、题解​​

​​3、源码​​

​​ 五、删除重复的电子邮箱​​

​​1、题目描述​​

​​2、题解​​

​​3、源码​​

一、组合两个表

1、题目描述

2、题解

3、源码

select FirstName,LastName,City,State from Person left join Address on Person.PersonId=Address.PersonId;

二、超过经理收入的员工

1、题目描述

2、题解

3、源码

select 
    e1.Name AS Employee
from 
    Employee as e1,
    Employee as e2
where
    e1.ManagerId = e2.Id AND e1.Salary>e2.Salary

 三、查找重复的电子邮箱

1、题目描述

2、题解

3、源码

方法一：
# select
#     Email
# from 
#     Person
# group by
#     Email
# having 
#     count(Email) > 1;

方法二：
select Email from
(
  select Email, count(Email) as num
  from Person
  group by Email
) as statistic
where num > 1
;

 四、从不订购的客户

1、题目描述

2、题解

3、源码

# Write your MySQL query 
select
    Name as Customers
from 
    Customers
where
    Customers.id not in 
(
    select customerid from orders
);

 五、删除重复的电子邮箱

1、题目描述

2、题解

3、源码

# Write your MySQL query statement below
DELETE p1 FROM Person p1,
    Person p2
WHERE
    p1.Email = p2.Email AND p1.Id > p2.Id