如何选择部门的前三名薪水？

Question

我正在尝试在线解决有2个表的SQL问题,如下所示,

员工表

Employee表包含所有员工.

+----+-------+--------+--------------+
| Id | Name  | Sa1ary | 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 |
+----+-------+--------+--------------+

部门表

SQL应该返回以下数据

我有以下SQL查询,

    SELECT D.Name AS Department, E.Name AS Employee, E.Salary AS Salary 
FROM Employee E INNER JOIN Department D ON E.DepartmentId = D.Id 
WHERE (SELECT COUNT(DISTINCT(Salary)) FROM Employee 
       WHERE DepartmentId = E.DepartmentId AND Salary > E.Salary) < 3
ORDER by E.DepartmentId, E.Salary DESC;

SQL很好,但是,我对这< 3部分感到困惑.它不应该是= 3,因为在被问及前3名工资的问题？我很感激SQL的解释,并会有所帮助.

Answer 1

如果您添加了一列, Count employees who earn more您的表格将如下所示

+----+-------+--------+--------------+-------------------------------+
| Id | Name  | Sa1ary | DepartmentId | Count employees who earn more |
+----+-------+--------+--------------+-------------------------------+
|  1 | Joe   |  70000 |            1 |    2                          |
|  2 | Henry |  80000 |            2 |    0                          |
|  3 | Sam   |  60000 |            2 |    1                          | 
|  4 | Max   |  90000 |            1 |    0                          |
|  5 | Janet |  69000 |            1 |    3                          |
|  6 | Randy |  85000 |            1 |    1                          |
+----+-------+--------+--------------+-------------------------------+

然后找到每个部门的前3名.你的WHERE会

WHERE `Count employees who earn more` < 3

如果你拥有=3它将返回唯一的第四高的员工

由于您没有该列,因此这就是SQL的作用

(SELECT COUNT(DISTINCT(Salary)) FROM Employee 
       WHERE DepartmentId = E.DepartmentId AND Salary > E.Salary)

如果您想生成上述表格,可以执行以下操作

SELECT 
      D.Name AS Department, 
      E.Name AS Employee, 
      E.Salary AS Salary,
      Count(E2.Salary) as Count_employees_who_earn_more
FROM Employee E 
INNER JOIN Department D 
ON E.DepartmentId = D.Id 
LEFT JOIN Employee E2 ON 
    e2.DepartmentId = E.DepartmentId
    AND E2.Salary > E.Salary
GROUP BY  D.Name , 
      E.Name , 
      E.Salary 

演示

Answer 2

我正在处理相同的 SQL 问题。

以防万一有人需要帮助。

这是我想出的答案。

SELECT
    dpt.Name AS Department,
    e1.Name AS Employee,
    e1.Salary AS Salary
FROM Employee AS e1
INNER JOIN Department dpt
ON e1.DepartmentID = dpt.Id
WHERE 3 > (
           SELECT COUNT(DISTINCT Salary)
           FROM Employee AS e2
           WHERE e2.Salary > e1.Salary
           AND e1.DepartmentID = e2.DepartmentID
          )
ORDER BY
Department ASC,
Salary DESC;

1. 难的是拿到每个部门的前三名。我先算一下【工资较高的员工人数】。

   之后，我使用3 > [工资较高的员工人数]只保留前 3 名的工资。（如果前3名员工超过3人，即部分工资相同，则全部计算在内。）

   询问

   SELECT *
   FROM Employee e1
   WHERE 3 > (
              SELECT COUNT(DISTINCT Salary)
              FROM Employee e2
              WHERE e2.Salary > e1.Salary
              AND e1.DepartmentID = e2.DepartmentID
             );
   

   输出

   +------+-------+--------+--------------+
   | Id   | Name  | Salary | DepartmentId |
   +------+-------+--------+--------------+
   |    1 | Joe   |  70000 |            1 |
   |    2 | Henry |  80000 |            2 |
   |    3 | Sam   |  60000 |            2 |
   |    4 | Max   |  90000 |            1 |
   |    6 | Randy |  85000 |            1 |
   +------+-------+--------+--------------+
   

2. 然后是容易的部分。您可以将这个表与 DepartmentID 上的 Department 连接起来以获取部门名称。

   最终输出

   +------------+----------+--------+
   | Department | Employee | Salary |
   +------------+----------+--------+
   | IT         | Max      |  90000 |
   | IT         | Randy    |  85000 |
   | IT         | Joe      |  70000 |
   | Sales      | Henry    |  80000 |
   | Sales      | Sam      |  60000 |
   +------------+----------+--------+