题解 | #查询各个岗位分数的中位数位置的范围#

想法1：考虑奇偶性

select m.job, 
       case when ct % 2 = 1 then ceiling(ct/2) else round(ct/2) end as start,
       case when ct % 2 = 1 then ceiling(ct/2) else round(ct/2+1) end as end
  from (
        select job, count(1) ct
          from grade
         group by job 
       ) m
 order by job;

想法2：不考虑奇偶性

select job,
       floor((count(job)+1)/2) as start,
       ceiling((count(job)+1)/2) as end
  from grade
 group by job
 order by job;