(202307)wonderful-sql:决胜秋招(task6)

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本文介绍了四个SQL练习，包括查询每个部门工资最高员工、座位调整、分数排名以及检测连续数字和至少有五名下属的经理。这些练习涵盖了数据库操作、数据统计和查询复杂度中等的内容。

教程链接：Datawhale - 一个热爱学习的社区

换硬盘重装了系统，后面应该也不会用到mysql，不装环境了，没有截图。

Section A

练习一: 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。

CREATE TABLE Employee (
    Id INT PRIMARY KEY,
    Name VARCHAR(255),
    Salary INT,
    DepartmentId INT,
    CONSTRAINT fk_department
        FOREIGN KEY (DepartmentId)
        REFERENCES Department(Id)
);

INSERT INTO Employee (Id, Name, Salary, DepartmentId) VALUES 
    (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。

CREATE TABLE Department (
    Id INT PRIMARY KEY,
    Name VARCHAR(255)
);

INSERT INTO Department (Id, Name) VALUES 
    (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。

SELECT d.Name AS Department, e.Name AS Employee, e.Salary
FROM Employee e
JOIN Department d ON e.DepartmentId = d.Id
WHERE (e.DepartmentId, e.Salary) IN (
  SELECT DepartmentId, MAX(Salary)
  FROM Employee
  GROUP BY DepartmentId
);

练习二: 换座位（难度：中等）

SELECT 
    CASE 
        WHEN id%2=0 THEN id-1 
        WHEN id<(SELECT COUNT(*) FROM seat) THEN id+1 
        ELSE id 
    END AS id, 
    student 
FROM seat 
ORDER BY id;

练习三: 分数排名（难度：中等)

SELECT 
    class, 
    score_avg, 
    RANK() OVER (ORDER BY score_avg DESC) AS rank1, 
    RANK() OVER (PARTITION BY score_avg ORDER BY class) AS rank2, 
    RANK() OVER (ORDER BY score_avg) AS rank3 
FROM (
    SELECT class, AVG(score) AS score_avg 
    FROM scores 
    GROUP BY class
) AS t
ORDER BY class;

练习四：连续出现的数字（难度：中等）

SELECT DISTINCT Num AS ConsecutiveNums
FROM (
    SELECT Num, 
           ROW_NUMBER() OVER (PARTITION BY Num ORDER BY Id) AS row_num, 
           ROW_NUMBER() OVER (PARTITION BY Num ORDER BY Id) - Id AS diff
    FROM Logs
) AS t
WHERE diff = 2
ORDER BY ConsecutiveNums;

练习六：至少有五名直接下属的经理（难度：中等）

SELECT Name
FROM Employee
WHERE ManagerId IS NULL
GROUP BY Name
HAVING COUNT(*) = 5;