查询各个岗位分数升序排列之后的中位数位置的范围floor()

/*1、case when语句，较为繁琐
with t as(
    select *,
        row_number() over(partition by job order by score) rk
    from grade
)

select job,
    case when mod(max(rk),2)=0 then round(max(rk)/2,0)
         when mod(max(rk),2)=1 then round((max(rk)+1)/2)
    end start,
    case when mod(max(rk),2)=0 then round(max(rk)/2+1)
         when mod(max(rk),2)=1 then round((max(rk)+1)/2)
    end end
from t
group by job
order by job

2、floor(x) 返回等于或小于x的最大整数；
ceil(x) 返回大于或等于x的最大整数
*/
select job,
    floor((count(score)+1)/2) as start,
    floor((count(score)+2)/2) as end
from grade
group by job
order by job