题解 | #考试分数(四)#

#中位数就要考虑job对应的数量的奇偶了

select
/*case 
    when mod(T1.num,2)=0 then num/2
    when mod(T1,num,2)<>0 then CEILING(T1.num)
end as start
#记错了语法，mod（除数，被除数)*/
T1.job1 as job,
case 
    when mod(T1.num,2)=0 then ROUND(T1.num/2,0)
    else CEILING(T1.num/2)
end as start
,case 
    when mod(T1.num,2)=0 then  ROUND(T1.num/2+1,0)
    else CEILING(T1.num/2)
end as end
from
(select
g.job as job1,
count(g.score) as num
from grade g
group by g.job) as T1
order by T1.job1

奇数，中位数[N/2]

偶数，中位数(N/2,N/2+1)