项目十：行程和用户（难度：困难）
Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘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 |
Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。
| 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 |
写一段 SQL 语句查出 2013年10月1日 至 2013年10月3日 期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。
| Day | Cancellation Rate |
|---|---|
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
作业代码：
-- 创建Trips表
CREATE TABLE Trips(
id INT PRIMARY KEY,
Client_id INT,
Driver_id INT,
City_id INT,
Status ENUM('completed','cancelled_by_driver','cancelled_by_client'),
Request_at VARCHAR(50)
);
-- 插入数据
INSERT INTO Trips VALUES ('1', '1', '10', '1', 'completed', '2013-10-01');
INSERT INTO Trips VALUES ('2', '2', '11', '1', 'cancelled_by_driver', '2013-10-01');
INSERT INTO Trips VALUES ('3', '3', '12', '6', 'completed', '2013-10-01');
INSERT INTO Trips VALUES ('4', '4', '13', '6', 'cancelled_by_client', '2013-10-01');
INSERT INTO Trips VALUES ('5', '1', '10', '1', 'completed', '2013-10-02');
INSERT INTO Trips VALUES ('6', '2', '11', '6', 'completed', '2013-10-02');
INSERT INTO Trips VALUES ('7', '3', '12', '6', 'completed', '2013-10-02');
INSERT INTO Trips VALUES ('8', '2', '12', '12', 'completed', '2013-10-03');
INSERT INTO Trips VALUES ('9', '3', '10', '12', 'completed', '2013-10-03');
INSERT INTO Trips VALUES ('10', '4', '13', '12', 'cancelled_by_driver', '2013-10-03');
-- 查看Trips表
SELECT * FROM Trips;
-- 创建Users表
CREATE TABLE Users(
Users_id INT PRIMARY KEY,
Banned varchar(20),
Role ENUM('client','driver','partner')
);
-- 插入数据
INSERT INTO Users VALUES ('1', 'No', 'client');
INSERT INTO Users VALUES ('2', 'Yes', 'client');
INSERT INTO Users VALUES ('3', 'No', 'client');
INSERT INTO Users VALUES ('4', 'No', 'client');
INSERT INTO Users VALUES ('10', 'No', 'driver');
INSERT INTO Users VALUES ('11', 'No', 'driver');
INSERT INTO Users VALUES ('12', 'No', 'driver');
INSERT INTO Users VALUES ('13', 'No', 'driver');
-- 查看Users表
SELECT * FROM users;
-- 作业解答
SELECT t.Request_at AS Day,
ROUND(sum((CASE WHEN t.Status LIKE 'cancelled%' THEN 1 ELSE 0 END))/count(*),2) AS 'Cancellation Rate' -- 如果是取消的就为1，否则为0，sum求和后除以当天的总单数，即为取消率。Round函数用来保留两位小数。
FROM Trips t
INNER JOIN Users u ON u.Users_Id =t.Client_Id AND u.Banned = 'No' -- 连接两张表
GROUP BY t.Request_at; -- 以订单时间分组
运行结果：
项目十一：各部门前3高工资的员工（难度：中等）
将昨天employee表清空，重新插入以下数据（其实是多插入5,6两行）：
| Id | Name | Salary | DepartmentId |
|---|---|---|---|
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
编写一个 SQL 查询，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：
| Department | Employee | Salary |
|---|---|---|
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
此外，请考虑实现各部门前N高工资的员工功能。
作业代码：要想实现各部门前N高只要替换where条件中的<3即可，前几就是小于几
-- 清空昨天的employee表
TRUNCATE TABLE employee;
SELECT *FROM employee
-- 插入数据
INSERT INTO employee VALUES (1,'Joe',70000,1);
INSERT INTO employee VALUES (2,'Henry',80000,2);
INSERT INTO employee VALUES (3,'Sam',60000,2);
INSERT INTO employee VALUES (4,'Max',90000,1);
INSERT INTO employee VALUES (5,'Janet',69000,1);
INSERT INTO employee VALUES (6,'Randy',85000,1);
-- 作业解答，
SELECT d.`Name` AS Department,e1.`Name` AS Employee,e1.Salary AS Salary
FROM employee e1
JOIN department d
ON e1.DepartmentId = d.id
WHERE (
SELECT COUNT(DISTINCT e2.Salary)
FROM employee e2
WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
) <3
ORDER BY d.`Name`,e1.Salary desc;
运行结果：
项目十二：分数排名（难度:中等）
依然是昨天的分数表，实现排名功能，但是排名是非连续的，如下：
| Score | Rank |
|---|---|
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 3 |
| 3.65 | 4 |
| 3.65 | 4 |
| 3.50 | 6 |
作业代码：
因为需要不连续的，所以count(*)取表中总记录数即可，之后统计多少个人比这个分数高，对结果+1后即为排名。因为比如对于最高分，没有人比他高，所以结果是0，排名需要+1才可以。并且使用format函数可以对结果强制保留几位小数输出。
SELECT FORMAT(Score,2),
(SELECT count(*) FROM Score AS s2 WHERE s2.Score > s1.Score)+1 AS Rank
FROM Score AS s1
ORDER BY Score DESC;
运行结果：
