[LeetCode] 262. 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 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
- [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 ...
- 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 ...
- 【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 ...
- 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 ...
- LeetCode (262):Nim Game
You are playing the following Nim Game with your friend: There is a heap of stones on the table, eac ...
- 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 ...
- 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 ...
- 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 ...
- 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
- 逆向破解之160个CrackMe —— 008-009
CrackMe —— 008 160 CrackMe 是比较适合新手学习逆向破解的CrackMe的一个集合一共160个待逆向破解的程序 CrackMe:它们都是一些公开给别人尝试破解的小程序,制作 c ...
- Linux centos通过安装lszrz用CRT实现与Windows互相传文件
本经验均在CentOSrelease6.7(Final)下操作,如知识有欠缺之处 欢迎批评指正: lrzsz是一个搭配SecureCRT使用的在linux和windows之间上传下载工具. 1 2 3 ...
- 2019-2020-1 20199301《Linux内核原理与分析》第九周作业
第八章 进程的切换和系统的一般执行过程 进程的调度实际与进程的切换 ntel定义的中断类型 硬中断:就是CPU的两根引脚(可屏蔽中断和不可屏蔽中断) 软中断/异常:包括除零错误.系统调用.调试断点等在 ...
- 【AirTest自学】AirTest工具介绍和入门学习(一)
==================================================================================================== ...
- 定时任务 Scheduled quartz
在项目应用中往往会用到任务定时器的功能,比如某某时间,或者多少多少秒然后执行某个骚操作等.spring 支持多种定时任务的实现,其中不乏自身提供的定时器.接下来介绍一下使用 spring 的定时器和使 ...
- SpringBoot终章(整合小型进销系统)
在前面的章节中我们学习Spring的时候可以看到配置文件比较多,所以我们有了SpringBoot 1. 引入依赖 <dependencies> <dependency> < ...
- [NOI2019]回家路线
[NOI2019]回家路线 题目大意: 有\(n\)个站点,\(m\)趟车,每趟车在\(p_i\)时从\(x_i\)出发,\(q_i\)时到达\(y_i\). 若小猫共乘坐了\(k\)班列车,依次乘坐 ...
- 洛谷P2331[SCOI2005]最大子矩阵
题目 DP 此题可以分为两个子问题. \(m\)等于\(1\): 原题目转化为求一行数列里的\(k\)块区间的和,区间可以为空的值. 直接定义状态\(dp[i][t]\)表示前i个数分为t块的最大值. ...
- 【洛谷】P4198 楼房重建(线段树)
传送门 分析 被线段树按在地上摩擦 先把左边转化成斜率,那么这个题就转化成每次修改一个点的值,输出前缀最大值的个数 看到标签是线段树,所以还是想想线段树的做法吧 既然是线段树,那么就要将区间分成两半 ...
- 用Visual Studio编写UDF的一点小技巧(二)