MySQL(II) 任务四

内容：

1. MySQL项目

1.1 数据的导入和导出

1.2 代码

2. 其他复杂项目

 

1. MySQL项目

1.1 数据的导入和导出

  以前文所创建的world表为例，导出到CSV文件到桌面的SQL文件夹，文件名为world.csv。

SELECT * FROM world
INTO OUTFILE '//Mac/Home/Desktop/SQL/world.csv'
FIELDS ENCLOSED BY '"'
TERMINATED BY ';'
ESCAPED BY '"'
LINES TERMINATED BY 'rn';

  导入的文件命名为world2。

LOAD DATA INFILE '//Mac/Home/Desktop/SQL/world.csv'
INTO TABLE world2
FIELDS TERMINATED BY ','
ENCLOSED BY '"'
LINES TERMINATED BY 'n'
IGNORE 1 ROWS;

 

1.2 代码

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。

CREATE TABLE Employee(
Id INT NOT NULL AUTO_INCREMENT ,
NAME VARCHAR(25) NOT NULL,
Salary INT NOT NULL,
DepartmentId INT NOT NULL,
PRIMARY KEY (Id)
);
INSERT INTO Employee (Name,Salary,DepartmentId) VALUES('Joe',70000,1),
('Henry',80000,2),
('Sam',60000,2),
('Max',90000,1);

CREATE TABLE Department(
Id INT NOT NULL AUTO_INCREMENT ,
NAME VARCHAR(25) NOT NULL,
PRIMARY KEY (Id)
);
INSERT INTO Department (Id,Name) VALUES(1,'IT'),
(2,'Sales'sample);

SELECT D.Name AS Department, E.NAME AS Employee,E.Salary
FROM employee E,
(SELECT DepartmentID, MAX(Salary) AS MAX FROM Employee GROUP BY DepartmentId) T,
Department D
WHERE E.DepartmentId=T.DepartmentId
AND E.Salary=T.MAX
AND E.DepartmentId=D.Id

 

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。

+---------+---------+
| 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。

SELECT @id:= @id +1 AS id, student FROM
(SELECT
case when MOD(id,2)=0 then id -1
when MOD(id,2) <>0 then id+1 AS id,
student FROM seat ORDER BY id) AS new_seat,
(SELECT @id :=0) AS init;

 

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 |
+-------+------+

SELECT Score, @rank:=@rank+(@prev <> (@prev := Score)) Rank
FROM Scores,
(SELECT @rank := 0, @prev := -1) init
ORDER BY Score DESC;

 

2. 复杂项目

项目十：行程和用户

SELECT Trips.request_at AS DAY,
ROUND(COUNT(case Trips.STATUS when 'Completed' then NULL ELSE 1 END) /COUNT(*),2)
AS 'Cancellation Rate'
FROM Users JOIN Trips
ON Users.Users_id=Trips.Client_id
WHERE Users.Banned ='No'
AND Trips.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY Trips.request_at；

 

项目十一：各部门前3高工资的员工

SELECT Department, Employee, Salary FROM(
SELECT Department, Employee, Salary, (case when @prevd = (@prevd := Department) AND @prevs <> (@prevs := salary)
then @c := @c+1 when @prevd =(@prevd : Department) AND @prevs =(@prevs := salary) then @c
ELSE @c :=1 END) AS c
FROM (
SELECT d.NAME AS Department, e.NAME AS Employee, Salary
FROM Employee AS e JOIN Department AS d
ON e.departmentid=d.id
ORDER BY d.NAME,salary DESC) AS t JOIN (SELECT @c :=1, @prevd :=0, @prevs :=0) AS init
)
AS t2 WHERE t2.c <=3;