1、题目名称
Trips and Users（出租车预约取消率）
2、题目地址
https://leetcode.com/problems/trips-and-users/
3、题目内容
表Trips内存储了出租车的接单数据，包括Id、Client_Id、Driver_Id、City_Id、Status、Request_at共计6列，其中Status列是个枚举列，包括completed（已完成）、cancelled_by_driver（由司机取消）、cancelled_by_client（由乘客取消）三类：
+----+-----------+-----------+---------+--------------------+----------+
| 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|
+----+-----------+-----------+---------+--------------------+----------+
表Users存储了所有的用户数据信息，每个用户有一个唯一的Users_Id，Role是一个枚举列，包括client（客户）、driver（司机）、partner（合作伙伴）三项：
+----------+--------+--------+
| 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 |
+----------+--------+--------+
写一个SQL语句，查询非禁止客户（Users表中Banned列为No的客户）在2013年10月1日至2013年10月3日间的单据取消率，结果为四舍五入后的两位有效数字。
4、初始化数据库脚本
在MySQL数据库中建立一个名为LEETCODE的数据库，用MySQL命令行中的source命令执行下面脚本：
-- 执行脚本前必须建立名为LEETCODE的DATABASE
USE LEETCODE;
DROP TABLE IF EXISTS Trips;
CREATE TABLE Trips (
Id INT NOT NULL PRIMARY KEY,
Client_Id INT,
Driver_Id INT,
City_Id INT,
Status ENUM('completed', 'cancelled_by_driver', 'cancelled_by_client'),
Request_at DATE
);
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (1, 1, 10, 1, 'completed', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (2, 2, 11, 1, 'cancelled_by_driver', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (3, 3, 12, 6, 'completed', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (4, 4, 13, 6, 'cancelled_by_client', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (5, 1, 10, 1, 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (6, 2, 11, 6, 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (7, 3, 12, 6, 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (8, 2, 12, 12, 'completed', '2013-10-03');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (9, 3, 10, 12, 'completed', '2013-10-03');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at)
VALUES (10, 4, 13, 12, 'cancelled_by_driver', '2013-10-03');
DROP TABLE IF EXISTS Users;
CREATE TABLE Users (
Users_Id INT NOT NULL PRIMARY KEY,
Banned VARCHAR(5),
Role ENUM('client', 'driver', 'partner')
);
INSERT INTO Users (Users_Id, Banned, Role) VALUES (1, 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (2, 'Yes', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (3, 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (4, 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (10, 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (11, 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (12, 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES (13, 'No', 'driver');
5、解题SQL1
我一开始想的办法比较复杂，第3-7行的SQL查出每日的未被禁止的客户提交的所有单据数量，第8-14行的SQL查出这些单据中以日为单位每日由未被禁止的客户取消的单据数量，在SELECT子句中计算取消率。
SELECT A.Request_at AS DAY,
ROUND(IFNULL(B.CANCELLED / A.TOTAL, 0), 2) AS 'Cancellation Rate'
FROM (SELECT COUNT(*) AS TOTAL, Request_at
FROM Trips T1, Users U1
WHERE T1.Client_Id = U1.Users_Id AND U1.Banned = 'No'
GROUP BY Request_at
HAVING COUNT(*)) AS A
LEFT JOIN (SELECT COUNT(*) AS CANCELLED, Request_at
FROM (SELECT DISTINCT T.Id, T.Request_at
FROM Trips T, Users U
WHERE T.Status <> 'completed' AND
T.Client_Id = U.Users_Id AND U.Banned = 'No') AS TMP
GROUP BY TMP.Request_at
HAVING COUNT(*)) AS B ON A.Request_at = B.Request_at
WHERE A.Request_at >= '2013-10-01' AND A.Request_at <= '2013-10-03';6、解题SQL2
下面的SQL是一个比较简单的方法，通过 LEFT JOIN 跨表查询，并统计状态不为completed的出租车订单。
SELECT Request_at DAY,
ROUND(SUM(IF(Status = 'completed', 0, 1)) / COUNT(*), 2) 'Cancellation Rate'
FROM Trips t
LEFT JOIN Users t1 ON t.Client_Id = t1.Users_Id
WHERE t1.Banned = 'No' AND Request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY t.Request_at;END
来源：oschina
链接：https://my.oschina.net/u/1425762/blog/496047