ORACLE SQL：显示平均工资最高的部门工资最低答案

【问题标题】：ORACLE SQL: Show the lowest salary in the department with the highest average salaryORACLE SQL：显示平均工资最高的部门工资最低
【发布时间】：2015-05-10 03:54:24
【问题描述】：

注意，这是一道作业题。

请显示平均工资为最高平均工资的部门的部门编号和最低工资。

这是我目前所拥有的，

SELECT DEPARTMENT_ID, MAX_AVG_SALARY
FROM
  (SELECT DEPARTMENT_ID, AVG(SALARY) AS MAX_AVG_SALARY
  FROM EMPLOYEES
  GROUP BY DEPARTMENT_ID)
WHERE MAX_AVG_SALARY =
  (SELECT MAX(MAX_AVG_SALARY)
  FROM
    (SELECT DEPARTMENT_ID,
      AVG(SALARY) AS MAX_AVG_SALARY
    FROM EMPLOYEES
    GROUP BY DEPARTMENT_ID
    ));

我可以得到工资最高的department_id，但是如何找到同一个部门的最低工资呢？

请帮忙！

谢谢！

【问题讨论】：

标签： sql oracle

【解决方案1】：

SELECT MINIMUM_SALARY,DEPARTMENT_ID
FROM
(
 SELECT AVG(SALARY) AS AVERAGE_SALARY,
        MIN(SALARY) AS MINIMUM_SALARY,
        DEPARTMENT_ID
  FROM EMPLOYEES
 GROUP BY DEPARTMENT_ID
)EMPLOYEE_AGGREGATED
WHERE
    AVERAGE_SALARY = (SELECT MAX(AVG(SALARY)) FROM EMPLOYEES GROUP BY DEPARTMENT_ID)

【讨论】：

【解决方案2】：

analytic functions 的解决方案：

select department_id, ms min_salary 
  from (
    select department_id, max(avg(salary)) over () mav,
        min(min(salary)) over (partition by department_id) ms, 
        min(avg(salary)) over (partition by department_id) av 
      from employees group by department_id )
  where av = mav order by department_id

SQLFiddle demo

【讨论】：

【解决方案3】：

您可以使用可以重复使用的命名子查询：

WITH grouped_salaries AS (
  SELECT department_id,
         MIN( salary ) AS min_dept_salary,
         AVG( salary ) AS avg_dept_salary
  FROM   Employees
  GROUP BY department_id
)
SELECT department_id,
       min_dept_salary
FROM   grouped_salaries
WHERE  avg_dept_salary = ( SELECT MAX( avg_dept_salary )
                           FROM   grouped_salaries );

【讨论】：

【解决方案4】：

我会使用分析函数来解决这个问题，特别是row_number() 来获得平均工资最高的部门：

select department_id, mins 
from (select department_id, avg(salary) as avgs, min(salary) as mins,
             row_number() over (order by avg(salary) desc) as seqnum
      from employees
      group by department_id
    ) de
where seqnum = 1;

【讨论】：

【解决方案5】：

select e.department_id, min(e.salary) min
  from employees e
having avg(e.salary) = (select max(avg(e.salary))
                          from employees e
                         group by e.department_id)
 group by e.department_id

【讨论】：

请编辑您的问题以添加解释。见how to answer。

【解决方案6】：

WITH CTE(ID, SAL) AS 
(
   SELECT DISTINCT
      DEPARTMENT_ID,
      AVG(SALARY) OVER(PARTITION BY DEPARTMENT_ID)"AVGSAL" 
   FROM
      EMPLOYEES 
   ORDER BY
      AVGSAL DESC
)
SELECT
   MIN(SALARY) 
FROM
   EMPLOYEES E1 
WHERE
   E1.DEPARTMENT_ID = 
   (
      SELECT
         ID 
      FROM
         CTE 
      WHERE
         ROWNUM = 1
   )
;

【讨论】：

请编辑您的问题以添加解释。见how to answer。

【解决方案7】：

这是不使用行限制功能的答案，因为旧的 oracle 不提供行限制功能

SELECT job_id, avg(salary) FROM Employees 
GROUP BY job_id 
HAVING AVG(salary) = (SELECT MIN(AVG(salary)) FROM EMPLOYEES GROUP BY job_id);

【讨论】：