select job,floor((count(*)+1)/2) start,ceil((count(*)+1)/2) end from grade group by job order by job
思路:求中位数的上下位注意在原数值基础上加一再除二,然后利用floor和ceil函数进行向下和向上取整。
select job,floor((count(*)+1)/2) start,ceil((count(*)+1)/2) end from grade group by job order by job
思路:求中位数的上下位注意在原数值基础上加一再除二,然后利用floor和ceil函数进行向下和向上取整。