sql练习三（DataWhale 系列-最终）

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
• 再将CSV表导入数据库

作业

项目七: 各部门工资最高的员工（难度：中等）

答案

---建表---
CREATE TABLE Employee(Id INT,
        NAME VARCHAR(10),
        Salary INT,
        DepartmentId INT); 
CREATE TABLE Department(Id INT,
        NAME VARCHAR(10));
---插值---
INSERT INTO Employee VALUES (1,'Joe',70000,1),
            (2,'Henry',80000,2),
            (3,'Sam',60000,2),
            (4,'Max',90000,1);
INSERT INTO Department VALUES(1,'IT'),(2,'Sales');
---查找---
---查找部门最高工资---
SELECT d.Name AS Department,e.name AS Employee,e.Salary FROM 
    Employee e LEFT JOIN department d ON e.DepartmentId=d.Id
    WHERE e.Salary=(SELECT MAX(Salary) FROM Employee WHERE DepartmentId=d.Id);
---查找部门工资最高前三---
SELECT Department.Name AS Department, e1.Name AS Employee, e1.Salary AS Salary
FROM Employee e1
JOIN Department
ON e1.DepartmentId = Department.Id
WHERE 3 > (
SELECT COUNT(DISTINCT e2.Salary) 
FROM Employee e2
WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
)
ORDER BY Department.Name, e1.Salary DESC

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：

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。

---建表插值---
CREATE TABLE seat(id INT,Student VARCHAR(10));
INSERT INTO seat VALUES(1,'Abbot'),
            (2,'Doris'),
            (3,'Emerson'),
            (4,'Green'),
            (5,'Jeames');
---转换座位---
SELECT(    CASE
    WHEN id %2 = 1 AND id!=max_id THEN id+1
    WHEN id %2 = 0 THEN id-1
    WHEN id = max_id THEN id
    END) AS id,student 
    FROM(SELECT id, student, (SELECT MAX(id) FROM seat) AS max_id FROM seat) a
    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”。

答案

---建表，插值---
CREATE TABLE Scores (Id INT,Score DOUBLE);
INSERT INTO Scores VALUES(1,3.50),(2,3.65),(3,4.00),
            (4,3.85),(5,4.00),(6,3.65);
----查询--
SELECT u.score,a.rank AS Rank FROM Scores u,
(SELECT @counter:=@counter+1 AS rank,t.score FROM (SELECT @counter:=0,score FROM Scores GROUP BY score ORDER BY score DESC)AS t)a
WHERE u.score=a.score ORDER BY Rank ASC;

4.2 MySQL 实战 - 复杂项目

作业

项目十：行程和用户（难度：困难）

答案

---创建插值---
CREATE TABLE Trips(Id INT,Client_Id INT,Driver_id INT,City_Id INT,STATUS VARCHAR(10),Request_at DATETIME);

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'),
        (1,4,13,12,'cancelled_by_driver','2013-10-03');

CREATE TABLE Users(Users_Id INT,Banned VARCHAR(3),Role VARCHAR(6));

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');

---查询---
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;

项目十一：各部门前3高工资的员工（难度：中等）

此外，请考虑实现各部门前N高工资的员工功能。

---查找前三---
SELECT Department.Name AS Department, e1.Name AS Employee, e1.Salary AS Salary
FROM Employee e1
JOIN Department
ON e1.DepartmentId = Department.Id
WHERE 3 > (
SELECT COUNT(DISTINCT e2.Salary) 
FROM Employee e2
WHERE e2.Salary > e1.Salary AND e1.DepartmentId = e2.DepartmentId
)
ORDER BY Department.Name, e1.Salary DESC

---查找前N的排名，把where后3改成N就好---

项目十二 分数排名 - （难度：中等）

---查询---
SELECT s.score,(SELECT COUNT(*)+1 FROM scores AS s1 WHERE                     s1.score>s.score) AS rank
    FROM scores s ORDER BY score DESC;