关于中位数的推论:当正序和逆序均大于总数的一半(向下取整)时,即为中位数
select grade from (
select
grade,
(select sum(number) from class_grade) as total ,
sum(number) over(
order by
grade asc
) as rank1,
sum(number) over(
order by
grade desc
) as rank2
from
class_grade
) t1
where rank1 >= total / 2 and rank2 >= total / 2
order by 1 asc