1 任务
任务时间
请于3月4日22:00前完成，在本文章评论打卡。逾期尚未打卡的会被清退。
作业
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 |
+------------+-------------------+
赠送创建表格代码，开不开心，见附件 :-)
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。
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 |
+-------+------+
2 作业
- 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 |
±-----------±------------------+
解答：
-- 创建表
CREATE TABLE IF NOT EXISTS Trips (
Id INT,
Client_Id INT,
Driver_Id INT,
City_Id INT,
Status ENUM('completed', 'cancelled_by_driver', 'cancelled_by_client'),
Request_at VARCHAR(50)
);
CREATE TABLE IF NOT EXISTS Users (
Users_Id INT,
Banned VARCHAR(50),
Role ENUM('client', 'driver', 'partner')
);
-- 表格中的资料会完全消失，可是表格本身会继续存在。
TRUNCATE TABLE Trips;
-- 插入数据
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');
-- 表格中的资料会完全消失，可是表格本身会继续存在。
TRUNCATE TABLE Users;
-- 插入数据
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');
-- 查询
SELECT t.Request_at AS 'Day',
ROUND((SUM(CASE WHEN t.Status LIKE 'cancelled%' THEN 1 ELSE 0 END))/COUNT(*),2) AS 'Cancellation Rate' -- 用ROUND()保留两位小数，用like通配符进行过滤
FROM Trips AS t INNER JOIN Users AS u
ON u.Users_Id = t.Client_Id AND u.Banned = 'No' -- 链接两个表
GROUP BY t.Request_at ; -- 以订单时间作为分组运行结果：
2. 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。
解答：
-- 创建表
CREATE TABLE Employee(
Id INT NOT NULL,
Name VARCHAR(20) NOT NULL,
Salary INT NOT NULL,
DepartmentId INT NOT NULL
);
CREATE TABLE Department(
Id INT NOT NULL,
Name VARCHAR(20) NOT NULL
);
-- 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);
INSERT INTO Department VALUES(1, 'IT');
INSERT INTO Department VALUES(2, 'Sales');
-- 查询
SELECT d.Name as Department, e.Name as Employee, e.salary as Salary
FROM Employee e INNER JOIN Department D ON e.DepartmentId = d.Id
WHERE(
SELECT COUNT(DISTINCT e2.Salary)
FROM Employee e2
WHERE e2.Salary > e.Salary AND e.DepartmentId = e2.DepartmentId -- e2.Salary > e.Salary 就是把表e里的序号一个个取出来，然后分别和整张表e2比较。只要需要一个值去扫描一张表的思路的地方就都可以用
)<3
ORDER BY d.Name, e.Salary DESC;运行结果：
3. 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 |
±------±-----+
解答：
DROP TABLE score;
-- 创建表
CREATE TABLE score(
Id INT NOT NULL,
Score FLOAT NOT NULL
);
-- 插入数据
INSERT INTO score VALUES(1, 3.50);
INSERT INTO score VALUES(2, 3.65);
INSERT INTO score VALUES(3, 4.00);
INSERT INTO score VALUES(4, 3.85);
INSERT INTO score VALUES(5, 4.00);
INSERT INTO score VALUES(6, 3.65);
-- 查询
SELECT score, (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：
