Question

在嵌套聚合函数中，使用group by子句ex max(avg (salary ))计算每个部门的第一个工资，假设分组由dept完成，然后选择max。现在我的理解是多个聚合函数如何工作

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

这样的查询将如何运作？哪个聚合函数首先执行？任何人都可以帮助我，我是SQL新手。

Answer 1

假设你有这张表：

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);

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department_id（30）的最大值和每个部门的平均工资的最大值。这些值无关。换句话说，您的SQL显示以前输出的最大值。

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，您可以选择多种方式如果有多个这样的部门，一切都取决于你想做什么，就像在我的例子中，两个（10和20）具有相同，最高的平均值：

-- 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}

Oracle sql嵌套聚合函数

1 个答案: