#每个岗位分数前2名用户 按语言名称升序+积分降序+grade的id升序 select p.id ,l.name ,p.score from ( select id ,language_id ,score ,dense_rank()over(partition by language_id order by score desc) r from grade#岗位分数排名 ) p left join language l on l.id=p.language_id where r<3#岗位分数排名前2名 order by 2,3 desc,1
