MySQL 4
4.1 MySQL 实战
感谢 https://blog.csdn.net/u011490595/article/details/82050325
让我终于连接上了navicat，终于不用看性冷淡的命令行了。
#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
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.
创建 Department 表，包含公司所有部门的信息。
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。（也可以使用RANK函数）
SELECT d. NAME AS Department, e. NAME AS Employee, e.Salary FROM Employee e
JOIN (SELECT Max(Salary) AS Salary FROM Employee GROUP BY DepartmentId)
AS g ON e.Salary = g.Salary LEFT JOIN Department d ON 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。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 (CASE WHEN MOD(id,2)=1 AND id != (SELECT COUNT(*)FROM Seat) THEN id+1
WHEN MOD(id,2)=0 THEN id-1 ELSE id END) AS id, student From Seat
ORDER BY id ASC;
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”。
SELECT Score,(SELECT COUNT(*) FROM (SELECT DISTINCT Score S FROM score) AS s2
WHERE S >= Score) AS 'Rank' FROM score ORDER BY score DESC;
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’)。 enum可以用来约束输入的值。
CREATE TABLE Trips (Id INT auto_increment 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_all DATE NOT NULL);
INSERT INTO Trips VALUES (NULL,1,10,1,'completed','2013-10-1');
INSERT INTO Trips VALUES (NULL,2,11,1,'cancelled_by_driver','2013-10-01'),
(NULL,3,12,6,'completed','2013-10-01'),(NULL,4,13,6,'cancelled_by_client','2013-10-01'),(NULL,1,10,1,'completed','2013-10-02'),
(NULL,2,11,6,'completed','2013-10-02'),(NULL,3,12,6,'completed','2013-10-02'),
(NULL,2,12,12,'completed','2013-10-03'),(NULL,3,10,12,'completed','2013-10-03'),(NULL,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。
CREATE TABLE Users (Users_Id INT NOT NULL PRIMARY KEY, Banned enum('Yes','No'), Role enum('client','driver','partner'));
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');
SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM
(SELECT Name,Salary, DepartmentId,
RANK() OVER (PARTITION BY DepartmentId ORDER BY Salary DESC) AS 'rank'
FROM Employee) AS e
LEFT JOIN Department d ON e.DepartmentId=d.Id WHERE e.rank<=3;
SELECT t.Request_all AS Day, ROUND(SUM(CASE WHEN Status ='completed' THEN 0 ELSE 1 END)/COUNT(*),2) AS 'Cancellation Rate'
FROM Trips t LEFT JOIN Users U ON t.Client_Id=U.Users_Id LEFT JOIN Users U2 on t.Driver_Id=U2.Users_Id
where U.Banned='No' AND U2.Banned='No' AND t.Request_all BETWEEN '2013-10-01'and'2013-10-03' GROUP BY t.Request_all;
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。
SELECT d.Name AS Department, e.Name AS Employee, e.Salary FROM (SELECT Name,Salary, DepartmentId, RANK() OVER (PARTITION BY DepartmentId ORDER BY Salary DESC) AS 'rank' FROM Employee) AS e LEFT JOIN Department d ON e.DepartmentId=d.Id WHERE e.rank<=3
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
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 3 |
| 3.65 | 4 |
| 3.65 | 4 |
| 3.50 | 6 |
+-------+------
在这里可以用Rank函数，也可以用上面的方式。
SELECT Score, RANK() OVER (ORDER BY Score DESC) AS 'Rank' FROM score
