• leetcode刷题笔记一百八十一题 && 一百八十二题 && 一百八十三题


    leetcode刷题笔记一百八十一题 && 一百八十二题 && 一百八十三题

    源地址:

    181. 超过经理收入的员工

    182. 查找重复的电子邮箱

    183. 从不订购的客户

    181问题描述:

    Employee 表包含所有员工,他们的经理也属于员工。每个员工都有一个 Id,此外还有一列对应员工的经理的 Id。

    +----+-------+--------+-----------+
    | Id | Name | Salary | ManagerId |
    +----+-------+--------+-----------+
    | 1 | Joe | 70000 | 3 |
    | 2 | Henry | 80000 | 4 |
    | 3 | Sam | 60000 | NULL |
    | 4 | Max | 90000 | NULL |
    +----+-------+--------+-----------+
    给定 Employee 表,编写一个 SQL 查询,该查询可以获取收入超过他们经理的员工的姓名。在上面的表格中,Joe 是唯一一个收入超过他的经理的员工。

    +----------+
    | Employee |
    +----------+
    | Joe |
    +----------+

    # Write your MySQL query statement below
    SELECT a.Name AS Employee FROM Employee a JOIN Employee b ON a.ManagerId = b.Id WHERE a.Salary > b.Salary;
    

    182问题描述:

    编写一个 SQL 查询,查找 Person 表中所有重复的电子邮箱。

    示例:

    +----+---------+
    | Id | Email |
    +----+---------+
    | 1 | a@b.com |
    | 2 | c@d.com |
    | 3 | a@b.com |
    +----+---------+
    根据以上输入,你的查询应返回以下结果:

    +---------+
    | Email |
    +---------+
    | a@b.com |
    +---------+
    说明:所有电子邮箱都是小写字母。

    # Write your MySQL query statement below
    SELECT Email FROM (
        SELECT Email, COUNT(Email) AS counte FROM Person  GROUP BY Email
    ) AS TempTable WHERE counte > 1 ;
    

    183问题描述:

    某网站包含两个表,Customers 表和 Orders 表。编写一个 SQL 查询,找出所有从不订购任何东西的客户。

    Customers 表:

    +----+-------+
    | Id | Name |
    +----+-------+
    | 1 | Joe |
    | 2 | Henry |
    | 3 | Sam |
    | 4 | Max |
    +----+-------+
    Orders 表:

    +----+------------+
    | Id | CustomerId |
    +----+------------+
    | 1 | 3 |
    | 2 | 1 |
    +----+------------+
    例如给定上述表格,你的查询应返回:

    +-----------+
    | Customers |
    +-----------+
    | Henry |
    | Max |
    +-----------+

    # Write your MySQL query statement below
    SELECT a.Name AS 'Customers' FROM Customers a WHERE a.Id NOT IN (SELECT CustomerId FROM Orders);
    
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  • 原文地址:https://www.cnblogs.com/ganshuoos/p/13656412.html
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