【数据库】MySQL 实战 - 复杂项目
4.2 MySQL 实战 - 复杂项目
#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 |
+------------+-------------------+
项目十一：各部门前3高工资的员工（难度：中等）
将项目7中的 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 |
+-------+------
4.2 MySQL 实战 - 复杂项目
项目十：行程和用户（难度：困难）（还没写出来，参考了别人的）
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：
1、创建Users表，并为它插入数据：
CREATE TABLE Users (
Users_Id INT NOT NULL PRIMARY KEY,
Banned varchar(3) NOT NULL,
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');
2、创建Trip表，并为它插入数据：
CREATE TABLE Trips (
Id INT NOT NULL PRIMARY KEY,
Client_Id INT NOT NULL,
Driver_Id INT NOT NULL,
City_Id INT NOT NULL,
STATUS ENUM('completed', 'cancelled_by_driver','cancelled_by_client'),
Request_at DATE,
FOREIGN KEY(Client_Id) REFERENCES Users(Users_Id),
FOREIGN KEY(Driver_Id) REFERENCES Users(Users_Id)
);
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');
3、查询;
SELECT T2.DAY,IFNULL(ROUND((T1.num/T2.num),2),0) AS 'Cancellation Rate'
FROM
(SELECT Request_at AS DAY,COUNT(*) AS num
FROM Trips t
LEFT JOIN Users u
ON t.Client_Id = u.Users_Id
WHERE u.Banned != 'Yes'
AND t.status != 'completed'
AND Request_at >='2013-10-01' AND Request_at <= '2013-10-03'
GROUP BY DAY) AS T1
RIGHT JOIN
(SELECT Request_at AS DAY,COUNT(*) AS num
FROM Trips t
LEFT JOIN Users u
ON t.Client_Id = u.Users_Id
WHERE u.Banned != 'Yes'
AND Request_at >='2013-10-01' AND Request_at <= '2013-10-03'
GROUP BY DAY) AS T2
ON T1.DAY = T2.DAY;
项目十一：各部门前3高工资的员工（难度：中等）
将项目7中的 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：
1、删除Employee表：
2、创建新的Employee表：
CREATE TABLE Employee (
id INT NOT NULL PRIMARY KEY,
name VARCHAR(255),
salary INT,
departmentid INT
);
2、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);
3、查询数据：
SELECT
d. NAME AS Department,
e. NAME AS Employee,
e.salary AS Salary
FROM employee AS e
INNER JOIN department AS d ON e.DepartmentId = d.id
WHERE (
SELECT COUNT(DISTINCT salary)
FROM employee
WHERE salary > e.salary
AND departmentid = e.DepartmentId
) < 3
ORDER BY e.departmentid,Salary DESC
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：
1、创建scores表：
CREATE TABLE scores (
Id INT NOT NULL PRIMARY KEY,
Score FLOAT(5,2) NOT NULL
);
2、插入数据：
INSERT INTO scores(Id,Score) VALUES(1,3.50);
INSERT INTO scores(Id,Score) VALUES(2,3.65);
INSERT INTO scores(Id,Score) VALUES(3,4.00);
INSERT INTO scores(Id,Score) VALUES(4,3.85);
INSERT INTO scores(Id,Score) VALUES(5,4.00);
INSERT INTO scores(Id,Score) VALUES(6,3.65);
3、查询数据：
SELECT
s.Score,
(SELECT
COUNT(*) + 1
FROM
Scores AS s1
WHERE
s1.Score > s.Score) AS Rank
FROM
scores s
ORDER BY Score DESC;