MySQL零基础入门---项目实战
MySQL入门基础
- 3.1 MySQL 实战
- 学习内容
- 作业
- 3.2 MySQL 实战 - 复杂项目
- 作业
3.1 MySQL 实战
学习内容
数据导入导出
- 将之前创建的任意一张MySQL表导出,且是CSV格式
- 再将CSV表导入数据库
作业
（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 departmentId。
+----+-------+--------+--------------+
| 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 |
+------------+----------+--------+
创建Employee 表，并填充数据：
CREATE TABLE Employee (
Id INT NOT NULL PRIMARY KEY,
Name VARCHAR(25) NOT NULL,
Salary INT NOT NULL,
DepartmentId INT NOT NULL
);
insert into Employee(Id,Name,Salary,DepartmentId) values(1,'Joe',70000,1);
insert into Employee(Id,Name,Salary,DepartmentId) values(2,'Henry',80000,2);
insert into Employee(Id,Name,Salary,DepartmentId) values(3,'Sam',60000,2);
insert into Employee(Id,Name,Salary,DepartmentId) values(4,'Max',90000,1);
创建Department 表，并填充数据：
CREATE TABLE Department (
Id INT NOT NULL PRIMARY KEY,
Name VARCHAR(25) NOT NULL
);
insert into Department(Id,Name) values(1,'IT');
insert into Department(Id,Name) values(2,'Sales');
编写一个 SQL 查询，找出每个部门工资最高的员工。
SELECT
d.Name AS Department, e.Employee, e.Salary
FROM
Department d
JOIN
(SELECT
Name AS Employee, Salary, DepartmentId
FROM
Employee
WHERE
Salary IN (SELECT
MAX(salary)
FROM
Employee
GROUP BY DepartmentId)) e ON d.Id = e.DepartmentId
ORDER BY Salary DESC;
（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。
创建seat 表，并填充数据：
CREATE TABLE seat (
id INT NOT NULL PRIMARY KEY,
student VARCHAR(25) NOT NULL
);
insert into seat(id,student) values(1,'Abbot');
insert into seat(id,student) values(2,'Doris');
insert into seat(id,student) values(3,'Emerson');
insert into seat(id,student) values(4,'Green');
insert into seat(id,student) values(5,'Jeames');
编写一个SQL 来改变相邻俩学生的座位。
SELECT (CASE
WHEN MOD(id,2) = 1 AND id = (SELECT COUNT(*) FROM seat) THEN id
WHEN MOD(id,2) = 1 THEN id+1
ElSE id-1
END) AS 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 |
+-------+------+
创建scores 表，并填充数据：
CREATE TABLE scores (
Id INT NOT NULL PRIMARY KEY,
Score float(5,2) NOT NULL
);
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);
编写一个 SQL 查询来实现分数排名。
SELECT
Score,
(SELECT
COUNT(DISTINCT Score)
FROM
scores
WHERE
Score >= s.Score) AS Rank
FROM
Scores s
ORDER BY Score DESC;
3.2 MySQL 实战 - 复杂项目
作业
（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 |
+------------+-------------------+
创建Users 表，并填充数据：
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');
创建Trips 表，并填充数据：
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');
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。
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;
（项目五、）各部门前3高工资的员工（难度：中等）
将项目一中的 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。
编写一个 SQL 查询，找出每个部门工资前三高的员工。
SELECT
d.name AS Department, e.Name AS Employee, Salary
FROM
Employee e
JOIN
Department d ON e.DepartmentId = d.Id
WHERE
(SELECT
COUNT(DISTINCT em.Salary)
FROM
Employee em
WHERE
em.Salary >= e.Salary
AND em.DepartmentId = e.DepartmentId) <= 3
GROUP BY Department , Salary DESC;
（项目六、）分数排名 - （难度：中等）
依旧是项目三中的分数表，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 |
+-------+------
用SQL 来实现排名功能，排名是非连续的。
SELECT
s.Score,
(SELECT
COUNT(*) + 1
FROM
Scores AS s1
WHERE
s1.Score > s.Score) AS Rank
FROM
scores s
ORDER BY Score DESC;