今天做的力扣SQL

发布于:2025-06-11 ⋅ 阅读:(18) ⋅ 点赞:(0)

我本地markdown的东西直接复制出来了。
多说一嘴,今天早上六点醒了,然后被外面吵,心里也担心找实习就一直睡不着了。索性直接来实验室,这一上午感觉好快啊。幸运的是,自己也没有浪费时间,还行吧。SQL欠的账迟早要还

181-employees-earning-more-than-their-managers

找出收入比经理高的员工

select a.name as 'Employee'
from employee as a, employee as b
where a.managerId = b.id and a.salary > b.salary

182-duplicate-emails

编写解决方案来报告所有重复的电子邮件。 请注意,可以保证电子邮件字段不为 NULL。

以 任意顺序 返回结果表。

查找重复的电子邮箱

select email
from Person
group by email
having count(email) > 1

183-customers-who-never-order

https://leetcode.com/problems/customers-who-never-order/description/
找出所有从不点任何东西的顾客。

从不订购的客户,2025年6月9日 星期一

select customers.name as 'Customers'
from customers
where Customers.id not in (select customerId from orders)

-- left join
select customers.name as 'Customers'
from customers
left join orders on Customers.id = Orders.customerId
where Orders.customerId is null

184-department-highest-salary

查找出每个部门中薪资最高的员工。
按 任意顺序 返回结果表。
https://leetcode.com/problems/department-highest-salary/description/

部门工资最高的员工

select Department.name as 'Department',
       Employee.name as 'Employee',
       Employee.salary as 'Salary'
from Employee
left join Department on Employee.departmentId = Department.id
where (Department.id, salary) in (
    select departmentId, max(salary)
    from Employee
    group by departmentId
    )