方法一：
select b.
*from
(select job,
floor((
count(*
)+1)
/2) as `start`,
floor((
count(*
)+1)
/2)+
if(count(*
)%2=
1,0,
1) as `end`
from grade group by job) a -- 中位數的位置資訊，也就是公升序之後的排名資訊
join
(select g1.*,
(select count
(distinct g2.score)
from grade g2
where g2.score>=g1.score and g1.job=g2.job) as t_rank
from grade g1 ) b -- 不同工作(job)裡面的每個人的資訊與排名
on(a.job=b.job and b.t_rank between a.start and a.end)
order by b.id
-- 要提取中位數的資訊，恰好a有中位數的資訊，那麼聯立a，b表，並且當工作(job)相同
--並且， b的排名(rank)在a表的中位數字置之間，那麼就表明這個資訊是中位數的資訊，最後聯立
select b.
*from
( select id,job,score,
row_number()
over
(partition by job order by score desc) as t_rank
from grade ) as b ,
(select job,
floor((
count
(id)+1
)/2) as mid_start,
floor((
count
(id)+2
)/2) as mid_end
from grade
group by job ) a
where b.job = a.job
and b.t_rank between a.mid_start and a.mid_end
order by b.id;
select a.id,a.job,a.score,b.t_rank
from
( select id,job,score,
row_number()
over
(partition by job order by score desc) as ranks
from grade ) a
join
( select job,
floor((
count
(score)+1
)/2) as t_rank
from grade
group by job
union
select job,
floor((
count
(score)/2
)+1) as t_rank
from grade
group by job ) b
on a.job = b.job
and a.ranks = b.t_rank
order by a.id;
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