4.1 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
关于csv文件，navicat for mysql中不包含cav文件的格式。尝试用MySQL workbench，可以导出csv文件。
再将CSV表导入数据库
使用sql语句：
导出
/*
字段之间以逗号分隔;字符串以半角双引号包围，字符串本身的双引号用两个双引号表示。
数据行之间以\r\n分隔；
*/
select * from test_info
into outfile 'test.csv'
fields terminated by ',' optionally enclosed by '"' escaped by '"'
lines terminated by '\r\n';
导入
load data infile 'test.csv'
into table test_info
fields terminated by ',' optionally enclosed by '"' escaped by '"'
lines terminated by '\r\n';
作业
项目七: 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。
USE test
CREATE TABLE IF NOT EXISTS Employee(
Id INT UNSIGNED AUTO_INCREMENT,
Name VARCHAR(100) NOT NULL,
Salary INT DEFAULT 0,
DepartmentId INT DEFAULT 0,
PRIMARY KEY (Id)
);
INSERT INTO Employee(Name,Salary,DepartmentId)
VALUES('Joe',70000,1),
('Henry',80000,2),
('Sam',60000,2),
('Max',90000,1);
USE test
CREATE TABLE IF NOT EXISTS Department(
Id INT UNSIGNED AUTO_INCREMENT,
Name VARCHAR(100) NOT NULL,
PRIMARY KEY (Id)
);
INSERT INTO Department(Name)
VALUES('IT'),
('Sales');
select d.Name as Department,e.Name as Employee,e.Salary
from Department d,Employee e
where e.DepartmentId=d.Id and e.Salary=(Select max(Salary) from Employee where DepartmentId=d.Id)
筛选结果
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
项目八: 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。
USE test
CREATE TABLE IF NOT EXISTS Seat(
id INT UNSIGNED AUTO_INCREMENT,
student VARCHAR(100) NOT NULL,
PRIMARY KEY (id)
);
INSERT INTO Seat(student)
VALUES('Adbot'),('Doris'),('Emerson'),('Green'),('James');
select s.id , s.student from
(
select id-1 as id ,student from seat where mod(id,2)=0
union
select id+1 as id,student from seat where mod(id,2)=1 and id !=(select count(*) from seat)
union
select id,student from seat where mod(id,2)=1 and id = (select count(*) from seat)
) s order by id;
项目九: 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 |
+-------+------+
sql语句
USE test
CREATE TABLE IF NOT EXISTS Scores(
Id INT UNSIGNED AUTO_INCREMENT,
Score FLOAT(5,2) NOT NULL,
PRIMARY KEY (Id)
);
INSERT INTO Scores(Score)
VALUES(3.50),(3.65),(4.00),(3.85),(4.00),(3.65);
select Score,
(select count(distinct Score) from Scores as s2 where s2.Score >= s1.Score) Rank
from Scores as s1
order by Score DESC;
4.2 MySQL 实战 - 复杂项目
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 |
+------------+-------------------+
sql语句
USE test
CREATE TABLE IF NOT EXISTS Users(
Users_Id INT DEFAULT 0,
Banned VARCHAR(100) NOT NULL,
Role ENUM('client','driver','partner') DEFAULT 'client'
);
INSERT INTO Users(Users_Id,Banned,Role)
VALUES(1,'NO','client'),(2,'Yes','client'),(3,'NO','client'),(4,'NO','client'),
(10,'NO','driver'),(11,'NO','driver'),(12,'NO','driver'),(13,'NO','driver');
SELECT Request_at Day,
ROUND(COUNT(IF(Status != 'completed', TRUE, NULL)) / COUNT(*), 2) 'Cancellation Rate'
FROM Trips
WHERE (Request_at between '2013-10-01' and '2013-10-03') and Client_Id IN (SELECT Users_Id FROM Users WHERE Banned = 'No')
GROUP BY Request_at;
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。
SELECT P2.Name AS Department,P3.Name AS Employee,P3.Salary AS Salary
FROM Employee AS P3
INNER JOIN Department AS P2
ON P2.Id = P3.DepartmentId
WHERE (
SELECT COUNT(DISTINCT Salary)
FROM Employee AS P4
WHERE P3.DepartmentId = P4.DepartmentId
AND P4.Salary >= P3.Salary
) <= 3
ORDER BY 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 |
sql语句
SELECT Score,
(SELECT COUNT(Score)
FROM score AS s2
WHERE s2.Score > s1.Score
)+1 AS rank
FROM score AS s1
ORDER BY Score DESC;