#对比前两条确实大巫见小巫，简单多了，前两条最怕是连题目都没读清
select level,count(distinct ui.uid) as level_cnt
from exam_record as er
join user_info as ui on er.uid=ui.uid
join examination_info as ei on er.exam_id=ei.exam_id
where ei.tag = "SQL"
and er.score >80
group by ui.level
order by level_cnt DESC,level DESC