# 思路1:将工资从小到大排序,奇数个,返回中间值,偶数个,返回中间两个值 # SELECT t.id,t.job,t.score,t.posn # from ( # SELECT id,job,score # ,row_number()over(partition by job order by score desc) posn # ,count(*)over(partition by job) cnt # from grade # ) as t # where posn in (floor(cnt/2)+1,if(mod(cnt,2)=0,floor(cnt/2),floor(cnt/2)...