[LeetCode] Customers Who Never Order 从未下单订购的顾客
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL query to find all customers who never order anything.
Table: Customers.
+----+-------+
| Id | Name |
+----+-------+
| 1 | Joe |
| 2 | Henry |
| 3 | Sam |
| 4 | Max |
+----+-------+
Table: Orders.
+----+------------+
| Id | CustomerId |
+----+------------+
| 1 | 3 |
| 2 | 1 |
+----+------------+
Using the above tables as example, return the following:
+-----------+
| Customers |
+-----------+
| Henry |
| Max |
+-----------+
这道题让我们给了我们一个Customers表和一个Orders表,让我们找到从来没有下单的顾客,那么我们最直接的方法就是找没有在Orders表中出现的顾客Id就行了,用Not in关键字,如下所示:
解法一:
SELECT Name AS Customers FROM Customers
WHERE Id NOT IN (SELECT CustomerId FROM Orders);
或者我们也可以用左交来联合两个表,只要找出右边的CustomerId为Null的顾客就是木有下单的:
解法二:
SELECT Name AS Customers FROM Customers
LEFT JOIN Orders ON Customers.Id = Orders.CustomerId
WHERE Orders.CustomerId IS NULL;
我们还可以用Not exists关键字来做,原理和Not in很像,参见代码如下:
解法三:
SELECT Name AS Customers FROM Customers c
WHERE NOT EXISTS (SELECT * FROM Orders o WHERE o.CustomerId = c.Id);
参考资料:
https://leetcode.com/discuss/22624/three-accepted-solutions
https://leetcode.com/discuss/53213/a-solution-using-not-in-and-another-one-using-left-join
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Customers Who Never Order 从未下单订购的顾客的更多相关文章
- LeetCode - Customers Who Never Order
Description: Suppose that a website contains two tables, the Customers table and the Orders table. W ...
- LeetCode——Customers Who Never Order(灵活使用NOT IN以及IN)
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- [SQL]LeetCode183. 从不订购的客户 | Customers Who Never Order
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- 183. Customers Who Never Order
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- [LeetCode]-DataBase-Customers Who Never Order
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- LeetCode:Binary Tree Level Order Traversal I II
LeetCode:Binary Tree Level Order Traversal Given a binary tree, return the level order traversal of ...
- LeeCode(Database)-Customers Who Never Order
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- 【SQL】183. Customers Who Never Order
Suppose that a website contains two tables, the Customers table and the Orders table. Write a SQL qu ...
- DataBase -- Customers Who Never Order
Question: Suppose that a website contains two tables, the Customers table and the Orders table. Writ ...
随机推荐
- [精品书单] C#/.NET 学习之路——从入门到放弃
C#/.NET 学习之路--从入门到放弃 此系列只包含 C#/CLR 学习,不包含应用框架(ASP.NET , WPF , WCF 等)及架构设计学习书籍和资料. C# 入门 <C# 本质论&g ...
- 再谈JavaScript闭包及应用
.title-bar { width: 80%; height: 35px; padding-left: 35px; color: white; line-height: 35px; font-siz ...
- jquery动态生成的元素添加事件的方法
动态生成的元素如果要添加事件,要写成 $(document).on("click", "#txtName", function() { alert(this.v ...
- 【无私分享:ASP.NET CORE 项目实战(第十四章)】图形验证码的实现
目录索引 [无私分享:ASP.NET CORE 项目实战]目录索引 简介 很长时间没有来更新博客了,一是,最近有些忙,二是,Core也是一直在摸索中,其实已经完成了一个框架了,并且正在准备在生产环境中 ...
- 数据结构:队列 链表,顺序表和循环顺序表实现(python版)
链表实现队列: 尾部 添加数据,效率为0(1) 头部 元素的删除和查看,效率也为0(1) 顺序表实现队列: 头部 添加数据,效率为0(n) 尾部 元素的删除和查看,效率也为0(1) 循环顺序表实现队列 ...
- MySQL的简单使用和JDBC示例
MySql是关系型数据库管理系统(RDBMS),所谓的"关系型"可以把它当作是"表格"概念,事实上,一个关系型数据库由一个或数个表格组成. MySQL所使用的S ...
- MVC Api 的跨项目路由
现有Momoda.Api项目,由于团队所有人在此项目下开发,导致耦合度太高,现从此接口项目中拆分出多个子项目从而避免对Momda.Api的改动导致“爆炸” MVCApi的跨项目路由和MVC有解决方式有 ...
- arcgis api for js入门开发系列三地图工具栏(含源代码)
上一篇实现了demo的地图加载展示,在上篇实现的基础上,新增了地图工具栏以及通用地图控件功能,比如地图框选缩放.地图漫游.清空.量算工具.地图导航控件.地图比例尺控件.地图鹰眼图等等,总共分为5个部分 ...
- iOS开发中常用的设计模式
常用的设计模式(一)代理模式应用场景:当一个类的某些功能需要由别的类来实现,但是又不确定具体会是哪个类实现.优势:解耦合敏捷原则:开放-封闭原则实例:tableview的 数据源delegate,通过 ...
- ios UIWebView自定义Alert风格的弹框
之前开发过一个App,因为公司之前写好了网页版的内容和安卓版本的App,我进去后老板要求我ios直接用网页的内容,而不需要自己再搭建框架.我一听,偷笑了,这不就是一个UIWebView吗?简单! 但是 ...