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【Mysql】基础 3

分类: 文章 • 2025-03-01 16:38:04

mysql实战

【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 |
±—±---------+

创建表：

create table Employee(
Id int auto_increment primary key,
Name varchar not null,
Salary int not null,
DepartmentId int not null
);
create table Department(
Id int primary key auto_increment,
Name varchar not null
);

插入数据：

insert into Employee (Name,Salary,DepartmentId) 
values
(Joe,70000,1),
(Henry,80000,2),
(Sam,60000,2),
(Max,90000,1);

insert into Department(Name) 
values (IT)，
	   (Sales);

找出每个部门工资最高的员工

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。
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。

创建表：

create table seat(
id int  not null auto_increment,
student varchar(15),
primary key (id)
);

插入数据：

insert into seat(student) 
values
	('Abbot'),
	('Doris'),
	('Emerson'),
	('Green'),
	('Jeames');

更新表格：

UPDATE seat s1
JOIN seat s2
ON (s1.id % 2 = 1 AND s2.id = s1.id+1)
SET s1.student=s2.student,s2.student=s1.student
WHERE s1.id+1 <> null;

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 |
±------±-----+

创建表：

create table score(
Id int auto_increment primary key,
Score float not null);

insert into score(Id,Score) 
values
	(1,3.50),
	(2,3.65),
	(3,4.00),
	(4,3.85),
	(5,4.00),
	(6,3.65);

SELECT Score,
(SELECT COUNT(*) 
FROM (SELECT DISTINCT Score S FROM score) AS s2

WHERE  S >= Score) AS 'Rank' 
FROM score ORDER BY score DESC;

【Mysql】基础 3

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’)。

CREATE TABLE Trips (
    Id INT NOT NULL 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_at date,
    foreign key(Client_Id) references Users(Users_Id),
    foreign key(Driver_Id) references Users(Users_Id)
);

CREATE TABLE Users (
    Users_Id INT NOT NULL PRIMARY KEY,
    Banned varchar(3) NOT NULL,
    Role ENUM('client', 'driver','partner')
);

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(Users_Id,Banned,Role) 
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');

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。

(SELECT Request_at Day,count(*) 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 between '2013-10-01' AND '2013-10-03'
	GROUP BY 1) AS T1
RIGHT JOIN
 (SELECT Request_at Day,count(*) num
	FROM Trips t
	LEFT JOIN Users u
	ON t.client_id = u.users_id
  WHERE u.banned != 'Yes'
  AND Request_at between '2013-10-01' AND '2013-10-03'
	GROUP BY 1) AS T2
  ON T1.DAY = T2.DAY

【Mysql】基础 3

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。

创建表


CREATE TABLE Employee (
id INT NOT NULL PRIMARY KEY,
name VARCHAR(255),
salary INT,
departmentid INT
);

CREATE TABLE Department (
id INT NOT NULL,
name varchar(255)
);

插入数据

INSERT INTO Employee VALUES
(1,'Joe',70000,1),
(2,'Henry',80000,2),
(3,'Sam',60000,2),
(4,'Max',90000,1),
(5,'Janet',69000,1),
(6,'Randy',85000,1);

INSERT INTO Department VALUES
(1,'IT'),
(2,'Sales')

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 AS Salary
FROM employee AS e
INNER JOIN department AS d ON e.DepartmentId = d.id
WHERE (
  SELECT count(DISTINCT salary)
	FROM employee
	WHERE salary > e.salary
	AND departmentid = e.DepartmentId
	) < 3
ORDER BY e.departmentid,Salary DESC

【Mysql】基础 3

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