MySQL项目实战

项目一: 各部门工资最高的员工（难度：中等）

创建Employee 表，包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。
±—±------±-------±-------------+
| Id | Name | Salary | DepartmentId |
±—±------±-------±-------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
±—±------±-------±-------------+
创建Department 表，包含公司所有部门的信息。
±—±---------+
| Id | Name |
±—±---------+
| 1 | IT |
| 2 | Sales |
±—±---------+
编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。
±-----------±---------±-------+
| Department | Employee | Salary |
±-----------±---------±-------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
±-----------±---------±-------+

CREATE TABLE Employee(
Id INT NOT NULL PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
Salary INT,
DepartmentId INT
);

INSERT INTO Employee (Id,Name,Salary,DepartmentId)
VALUES (1,'Joe',70000,1),(2,'Henry',80000,2),(3,'Sam',60000,2),(4,'Max',90000,2);

CREATE TABLE Department(
Id INT NOT NULL,
Name VARCHAR(255)
);


INSERT INTO Department (Id,Name)
VALUES (1,'IT'),(2,'Sales');

SELECT Department.Name As Department, Employee.Name As Employee, Employee.Salary As Salary
FROM Department JOIN Employee
ON Department.Id = Employee.DepartmentId
WHERE Salary IN (SELECT MAX(Salary) FROM Employee GROUP BY DepartmentId);

项目二: 换座位（难度：中等）

小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。
其中纵列的 id 是连续递增的
小美想改变相邻俩学生的座位。
你能不能帮她写一个 SQL query 来输出小美想要的结果呢？
请创建如下所示seat表：
示例：
±--------±--------+
| id | student |
±--------±--------+
| 1 | Abbot |
| 2 | Doris |
| 3 | Emerson |
| 4 | Green |
| 5 | Jeames |
±--------±--------+
假如数据输入的是上表，则输出结果如下：
±--------±--------+
| id | student |
±--------±--------+
| 1 | Doris |
| 2 | Abbot |
| 3 | Green |
| 4 | Emerson |
| 5 | Jeames |
±--------±--------+
注意：
如果学生人数是奇数，则不需要改变最后一个同学的座位。

CREATE TABLE seat(
id INT NOT NULL PRIMARY KEY,
student VARCHAR(255)
);

INSERT INTO seat
VALUES
(1,'Abbot'),
(2,'Doris'),
(3,'Emerson'),
(4,'Green'),
(5,'Jeames');


SELECT 
( CASE when MOD(id,2)=1 AND id!=(SELECT MAX(id) FROM seat )  then  id+1
       when MOD(ID,2)=0 then id-1
       ELSE id
       END) id ,student
 FROM seat
 ORDER BY id;

项目三: 分数排名（难度：中等）

编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。
创建以下score表：
±—±------+
| Id | Score |
±—±------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
±—±------+
例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：
±------±-----+
| Score | Rank |
±------±-----+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
±------±-----+

CREATE TABLE score(
Id INT,
Score DECIMAL(3,2)
);

INSERT INTO score
VALUES (1,3.50),(2,3.65),(3,4.00),(4,3.85),(5,4.00),(6,3.65);

SELECT Score,(
                SELECT COUNT(DISTINCT  Score) 
                FROM score 
                WHERE Score >= s.Score) AS 'Rank'
        FROM score AS s
ORDER BY Score DESC;
 

项目四：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 |
±-----------±------------------+

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)
);

CREATE TABLE Users (
Users_Id INT PRIMARY KEY, 
Banned VARCHAR(50), 
Role ENUM('client', 'driver', 'partner')
);


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');


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');

ALTER TABLE Trips
ADD FOREIGN KEY(Client_Id)
REFERENCES Users(Users_Id);

ALTER TABLE Trips
ADD FOREIGN KEY(Driver_Id)
REFERENCES Users(Users_Id);

SELECT Request_at as Day,
ROUND(SUM(CASE when Status  LIKE 'Cancelled%' then 1 else 0 END)/COUNT(*),2) as 'Cancellation Rate'
FROM Trips JOIN Users
ON Users.Users_Id = Trips.Client_Id AND Users.Banned = 'No'
GROUP BY Request_at;

项目五：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。

TRUNCATE TABLE employee;

INSERT INTO Employee
VALUES(1,'Joe',70000,1),(2,'Henry',80000,2),(3,'Sam',60000,2),
(4,'Max',90000,1),(5,'Janet',69000,1),(6,'Randy',85000,1);

SELECT d.Name as Department,e.Name as Employee,e.Salary as Salary
FROM Employee e 
JOIN Department d ON e.DepartmentId = d.Id
WHERE Salary IN(
                  SELECT Salary 
                  FROM Employee 
                  WHERE (SELECT COUNT(DISTINCT Salary) 
                  FROM Employee 
                  WHERE Salary >= e.Salary AND  DepartmentId=d.id) <=3)
ORDER BY DepartmentId,Salary DESC;

项目六 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 |
±------±-----

SELECT Score,((SELECT COUNT(Score) 
              FROM score 
              WHERE Score > s.Score) +1) as 'Rank'
FROM score as s
ORDER BY Score DESC;