两个聚合函数分组

Posted 2023-03-31

【中文标题】两个聚合函数分组

【英文标题】：Two aggregation functions group by

【发布时间】：2015-04-27 20:42:16

【问题描述】：

我正在尝试打印所有工资总和大于部门平均总和的部门名称。

SELECT d.department_name, SUM(e.salary)
FROM departments d, employees e
WHERE d.department_id = e.department_id
GROUP BY d.department_name
HAVING SUM(e.salary) > (SELECT AVG(SUM(salary)) from employees);

在第二次选择之后，我必须按AVG(SUM(salary)) 分组吗？

【问题讨论】：

查看使用窗口集（Over partition by），这可以让您通过选择获得平均内联。示例：oracle-base.com/articles/misc/analytic-functions.php 或者：***.com/questions/6847814/…

【参考方案1】：

您需要重复条件中的第一个查询。这可以通过 WITH 子句来完成。

WITH dept_sums AS (SELECT d.department_name, SUM(e.salary) sum_salary
FROM departments d, employees e
WHERE d.department_id = e.department_id
GROUP BY d.department_name)

SELECT * FROM dept_sums d_s_1 WHERE d_s_1.sum_salary > (SELECT AVG(sum_salary) FROM dept_sums d_s_2);

【参考方案2】：

这是窗口（分析）函数派上用场的地方。下面我使用AVG()作为解析函数来计算所有部门的平均总工资。

SELECT department_name, dept_salary FROM (
    SELECT d.department_name, SUM(e.salary) AS dept_salary
         , AVG(SUM(e.salary)) OVER ( ) AS avg_dept_salary
      FROM departments d INNER JOIN employees e
        ON d.department_id = e.department_id
     GROUP BY d.department_name
) WHERE dept_salary > avg_dept_salary;