# 用窗口函数统计出记录数量，再用distinct去重。

select distinct *
from
(
select d.dept_no,d.dept_name,count(s.salary) over (partition by d.dept_no) as sum
from departments d
     left join dept_emp de on d.dept_no = de.dept_no
     left join salaries s on de.emp_no = s.emp_no
) as t
order by dept_no