MySQL小项目练习

1、数据导入导出

1.1 导入sql

mysql> create database bookrecommend;
mysql> use bookrecommend;
mysql> set names utf8;
mysql> source /Users/bishounendai/Downloads/bookRecommend--2/数据库备份/bookrecomme

1.2导入csv文件

（1）现有一个csv文件test1.csv，放于桌面

（2）新建数据库和表

#新建数据库
mysql> create database test1;
mysql> use test1

#新建表
mysql> create table if not exists student(
    -> id int(11) not null,
    -> name varchar(45) not null,
    -> password varchar(45) not null,
    -> primary key (id)
    -> )ENGINE=InnoDB DEFAULT CHARSET=utf8;
#将csv文件导入mysql
mysql> load data infile '/Users/bishounendai/Desktop/test1.csv'--csv文件存放路径
    -> into table student--要将数据导入的表名
    -> fields terminated by ',' optionally enclosed by '"' escaped by '"'
    -> lines terminated by '\r\n';

遇到的坑：

The MySQL server is running with the --secure-file-priv option so it cannot execute this statement

解决方案：
https://www.jianshu.com/p/9ef06f5a58da
（3）查询插入是否成功：

mysql> select * from student;

1.3 导出数据

（1）将数据导出到text中

mysql> SELECT * FROM runoob_tbl 
    -> INTO OUTFILE '/tmp/runoob.txt';

（2）将数据导出到csv文件中

mysql> SELECT * FROM passwd INTO OUTFILE '/tmp/runoob.txt'
    -> FIELDS TERMINATED BY ',' ENCLOSED BY '"'
    -> LINES TERMINATED BY '\r\n';

2、作业

项目七：各部门工资最高的员工

任务：编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据下述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

(1)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。
CREATE TABLE Employee(
	Id INT(10) NOT NULL PRIMARY KEY,
	Name VARCHAR(20),
	Salary VARCHAR(20),
	DepartmentId INT(10)
);
(2)创建Department 表，包含公司所有部门的信息。
CREATE TABLE Department(
	Id INT(10) NOT NULL PRIMARY KEY,
	Name VARCHAR(20)
);
(3)插入数据
INSERT INTO Employee(
	Id,Name,Salary,DepartmentId
)VALUES(1,'Joe','70000',1),
		(2,'Henry','80000',2),
		(3,'Sam','60000',2),
		(4,'Max','90000',1);
INSERT INTO Department(
	Id,Name
)VALUES(1,'IT'),
		(2,'Sales');

展示数据表

#思路：根据DepartmentId进行左连接，员工表后加上一列行业，再根据DepartmentId进行分组，寻找每个行业中薪资最高的员工。
mysql> SELECT d.Name AS Department,
    -> e.Name AS Employee,
    -> e.Salary
    -> FROM Employee AS e
    -> LEFT JOIN 
    -> Department AS d ON e.DepartmentId = d.Id
    -> WHERE (e.DepartmentId, e.Salary) IN (SELECT e1.DepartmentId,MAX(e1.Salary) AS Salary FROM Employee e1 GROUP BY e1.DepartmentId);

项目八：换座位

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？
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。

(1)创建表
CREATE TABLE seat(
 	id INT(10) NOT NULL PRIMARY KEY, 
 	student VARCHAR(10) 
 ); 
 （2）插入数据
 INSERT INTO seat(id,student) 
VALUES
	(1,"Abbot"),
	(2,"Doris"),
	(3,"Emerson"),
	(4,"Green"),
	(5,"Jeames");
	（3）判断id 是奇数且最后一个 id不变。
	判断id是奇数且不为最后一个 id+1，判断id是偶数 id-1
SELECT (
		CASE WHEN id%2=1 and id = (SELECT COUNT(*) FROM seat) THEN id
		WHEN id%2 =1 THEN id+1
		ELSE id-1
		END
		) AS id,student
FROM seat
ORDER BY id;

结果如下：

项目九: 分数排名

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 |
±------±-----+

（1）创建表
CREATE TABLE score(
 	Id INT(10) NOT NULL PRIMARY KEY, 
 	Score FLOAT(4) 
 ); 
 （2）插入数据
 INSERT INTO score(Id,Score) 
VALUES
	(1,3.50),
	(2,3.65),
	(3,4.00),
	(4,3.85),
	(5,4.00),
	(6,3.65);

mysql> SELECT Score,(SELECT COUNT(DISTINCT score) from score WHERE score >= s.score) as Rank
		-> from score as s 
		-> ORDER BY Score DESC;

项目十：行程和用户

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 |
±-----------±------------------+

（1）创建表
CREATE TABLE trips(
    Id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
    Client_Id INT NOT NULL,
    Driver_Id INT NOT NULL,
    City_Id INT NOT NULL,
    Status ENUM('completed', 'cancelled_by_driver', 'cancelled_by_client') NOT NULL,
    Request_at DATE DEFAULT NULL);

CREATE TABLE Users(
    Users_Id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
    Banned VARCHAR(10) NOT NULL,
    Role ENUM('client', 'driver', 'partnet') NOT NULL);
（2）插入数据
INSERT INTO trips
     VALUES (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');

INSERT INTO Users
    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');
 （3）查询语句
 SELECT t.Request_at AS 'Day',
ROUND((SUM(CASE WHEN t.Status LIKE 'cancelled%' THEN 1 ELSE 0 END))/COUNT(*),2) AS 'Cancellation Rate' 
FROM Trips AS t INNER JOIN Users AS u 
ON u.Users_Id = t.Client_Id AND u.Banned = 'No' 
GROUP BY t.Request_at ;