Question

select count(*), department_id 
from employees 
group by department_id 
having count(*) > avg(count(*));

我只想选择>平均编号。

Answer 1

您可以在having子句中使用子查询：

select count(*), department_id 
from employees 
group by department_id 
having count(*) > (select avg(cnt)
                   from (select count(*) as cnt
                         from employees 
                         group by department_id
                        )
                  );

不需要额外级别的子查询。也可以写成：

select count(*), department_id 
from employees 
group by department_id 
having count(*) > (select count(*) / nullif(count(distinct department_id), 0)
                   from employees 
                  );

这只是通过将员工总数除以部门数来计算部门中的平均员工数。

Answer 2

您需要将平均值计算为：

select avg(counter) from (select count(*) as counter from employees group by department_id)

进行比较：

select 
  count(*), 
  department_id 
from employees 
group by department_id 
having 
  count(*) > (select avg(counter) from (select count(*) as counter from employees group by department_id));

Answer 3

您也可以将cross join用作

select e2.department_id
  from employees e2
 cross join (select count(e.empno) / count(distinct e.department_id) avg_emp
               from employees e) e1
 group by e2.department_id, avg_emp
having count(*) > e1.avg_emp;

Rextester Demo

如何解决计数（*）超过平均水平？

3 个答案: