sql练习三(DataWhale 系列-最终)

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sql练习三(DataWhale 系列-最终)

晓生寒 2019-04-08 15:48:22 浏览617
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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)

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。
Department Employee Salary
IT Max 90000
Sales Henry 80000

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:

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。

---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”。

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

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)

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

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)

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。

---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)

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
---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;

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