The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are both foreign keys to the Users_Id at the Users table. Status is an ENUM type of (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’).

+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id | Status |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1 | 1 | 10 | 1 | completed |2013-10-01|
| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01|
| 3 | 3 | 12 | 6 | completed |2013-10-01|
| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01|
| 5 | 1 | 10 | 1 | completed |2013-10-02|
| 6 | 2 | 11 | 6 | completed |2013-10-02|
| 7 | 3 | 12 | 6 | completed |2013-10-02|
| 8 | 2 | 12 | 12 | completed |2013-10-03|
| 9 | 3 | 10 | 12 | completed |2013-10-03|
| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+

The Users table holds all users. Each user has an unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’).

+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+

Write a SQL query to find the cancellation rate of requests made by unbanned clients between Oct 1, 2013 and Oct 3, 2013. For the above tables, your SQL query should return the following rows with the cancellation rate being rounded to two decimal places.

+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+

Trips表里有一些Id, 状态,请求时间。Users表里有顾客和司机信息, 还有该顾客和司机有没有被Ban的信息。要返回一个结果看某个时间段内由没有被ban的顾客提出的取消率是多少。其实题目没有说清楚顾客到底包不包括司机,其实是包括的,由司机提出的取消请求也应计算进去,用Case When ... Then ... Else ... End关键字来做,用cancelled%来表示开头是cancelled的所有项,这样就包括了driver和client,然后分母是所有项,限制条件里限定了时间段,然后是没有被ban的,结果需要保留两位小数,所以用Round关键字给定参数2。

解法1:

SELECT t.Request_at Day, ROUND(SUM(CASE WHEN t.Status LIKE 'cancelled%' THEN 1 ELSE 0 END)/COUNT(*), 2) 'Cancellation Rate'
FROM Trips t JOIN Users u ON t.Client_Id = u.Users_Id AND u.Banned = 'No'
WHERE t.Request_at BETWEEN '2013-10-01' AND '2013-10-03' GROUP BY t.Request_at;  

解法2:

SELECT Request_at Day, ROUND(COUNT(IF(Status != 'completed', TRUE, NULL)) / COUNT(*), 2) 'Cancellation Rate'
FROM Trips WHERE (Request_at BETWEEN '2013-10-01' AND '2013-10-03') AND Client_Id IN
(SELECT Users_Id FROM Users WHERE Banned = 'No') GROUP BY Request_at;

  

All LeetCode Questions List 题目汇总

[LeetCode] 262. Trips and Users 旅行和用户的更多相关文章

  1. [LeetCode] Trips and Users 旅行和用户

    The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are b ...

  2. leetcode——262. Trips and Users

    The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are b ...

  3. 【leetcode】Trips and Users

    The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are b ...

  4. 262. Trips and Users

    问题描述 解决方案 -- case when 的效率比if的效率高 -- select Trips.Request_at as 'Day', -- round(sum(case Trips.Statu ...

  5. LeetCode (262):Nim Game

    You are playing the following Nim Game with your friend: There is a heap of stones on the table, eac ...

  6. LeetCode All in One 题目讲解汇总(持续更新中...)

    终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance ...

  7. leetcode 数据库十题记录

    题目从难到易记录.解题过程中,如果不太熟悉,可以将题目中的表自己手动录入到自己的数据库中,就方便学习,测试. 185. Department Top Three Salaries 要求就是查询出每个部 ...

  8. All LeetCode Questions List 题目汇总

    All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems clas ...

  9. Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017)

    All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Las ...

随机推荐

  1. ADAS 实车道路及场地测试服务

    概述         高级驾驶员辅助系统(ADAS)作为车辆主动安全的关键系统,已经被越来越多的车辆配置.ADAS系统与车辆动力.制动.转向系统有直接耦合,是车辆中可靠性要求很高的控制系统,因此针对A ...

  2. moviepy改进的想码

    这个要比前一个厚实点. 更改视频亮度,增加字幕,去除音轨,淡入特效,转换,截取时间,控制位置,组合图框,合成多段, 嗯,很多都有了. from django.test import TestCase ...

  3. drf序列化器与反序列化

    什么是序列化与反序列化 """ 序列化:对象转换为字符串用于传输 反序列化:字符串转换为对象用于使用 """ drf序列化与反序列化 &qu ...

  4. python iter()函数迭代器

    迭代器为类序列对象提供了一个类序列的接口.python的迭代无缝地支持序列对象,而且它还允许程序员迭代非序列类型,包括用户定义的对象.迭代器用起来很灵巧,你可以迭代不是序列但表现处序列行为的对象,例如 ...

  5. HDU - 3535:AreYouBusy (分组背包)

    题意:给你n个工作集合,给你T的时间去做它们.给你m和s,说明这个工作集合有m件事可以做,它们是s类的工作集合(s=0,1,2,s=0说明这m件事中最少得做一件,s=1说明这m件事中最多只能做一件,s ...

  6. 手机代理调试Charles Proxy和Fiddler

    一.Charles Proxy Charles是一个HTTP代理/HTTP监控/反向代理的工具. 使用它开发者可以查看设备的HTTP和SSL/HTTPS网络请求.返回.HTTP头信息 (cookies ...

  7. SpringBoot源码分析-编译环境与新建测试模块

    建议 分析源码建议不要使用Idea或者Eclipse等IDE工具的反编译功能或者导入源码包的方式看源码,那样不能给框架的源码做注释,所以分析源码之前都得先下载源码并构建,然后在项目中新建一个Modul ...

  8. [Codeforces 1242C]Sum Balance

    Description 题库链接 给你 \(k\) 个盒子,第 \(i\) 个盒子中有 \(n_i\) 个数,第 \(j\) 个数为 \(x_{i,j}\).现在让你进行 \(k\) 次操作,第 \( ...

  9. vue 的computed 和 watch 两者的区别

    computed是计算属性,依赖其他属性计算,并且computed的值有缓存,只有当计算值发生变化才会返回内容. computed 用来监控自己定义的变量,该变量不在data里面声明,直接在compu ...

  10. 一个Tomcat下部署多个项目异常:org.springframework.jmx.export.UnableToRegisterMBeanException: Unable to register MBean 的解决方法

    内容简介 在测试服务器上Tomcat下部署两个Spring boot项目,总是一个能启动成功,另一个启动不成功.这两个war包单独部署均能正常启动. 查看日志:启动时报出 org.springfram ...