sql - Oracle sql 嵌套聚合函数

Question

在使用 group by 子句 ex max(avg (salary)) 的嵌套聚合函数中，假设分组是按部门完成的，则计算每个部门的第一个工资，然后选择其中的最大值。现在我不明白的是多个聚合函数是如何工作的

 select max(dept), max(avg(salary))
 from employee
 group by dept; 

这样的查询如何工作？哪个聚合函数首先执行？任何人都可以帮助我吗，我是 SQL 新手。

Answer 1

假设您有这张表:

create table employee (dept number(4), salary number(12));
insert into employee values (10, 1000);
insert into employee values (10, 5000);
insert into employee values (20, 2500);
insert into employee values (20, 3500);
insert into employee values (30, 2000);

所以各部门的平均工资是:

select dept, avg(salary) avs from employee group by dept order by dept;
DEPT        AVS
----- ----------
   10       3000
   20       3000
   30       2000

您的查询显示了department_id的最大值(30)和每个部门平均工资的最大值。这些值不相关。 换句话说，您的 SQL 显示了先前输出的最大值。

select max(dept), max(avg(salary)) from employee group by dept;
DEPT        AVS
----- ----------
   30       3000

---

如果您想显示平均工资最高的部门，您有多种选择。
如果有多个这样的部门，一切都取决于您想要做什么，就像在我的示例中，其中两个(10 和 20)具有相等的最高平均值:

-- use HAVING clause with subquery (old school):
select dept, avg(salary) avs
  from employee group by dept
  having avg(salary) = (select max(avg(salary)) from employee group by dept)
  order by dept;

-- use analytic MAX() function:
select dept, avs
  from (
    select dept, avg(salary) avs, max(avg(salary)) over () max_avs
      from employee group by dept)
  where avs = max_avs order by dept;

-- use analytic DENSE_RANK():
select dept, avs 
  from (
    select dense_rank() over (order by avg(salary) desc) rnk, dept, avg(salary) avs 
      from employee group by dept)
  where rnk = 1 order by dept;

-- if you want ONLY ONE ROW use MIN(...) or MAX(...) KEEP (DENSE_RANK...)
-- this returns department with lower ID:
select min(dept) keep (dense_rank last order by avg(salary)) dept,
       min(avg(salary)) keep (dense_rank first order by avg(salary) desc) avs
  from employee group by dept;

-- this returns department with higher ID:
select max(dept) keep (dense_rank last order by avg(salary)) dept,
       max(avg(salary)) keep (dense_rank first order by avg(salary) desc) avs
  from employee group by dept;

^{SQLFiddle demo}