题解 | #作答试卷得分大于过80的人的用户等级分布#

select t3.level as level, count(distinct t3.uid) as level_cnt
from exam_record as t1
inner join examination_info as t2 on t1.exam_id = t2.exam_id
inner join user_info as t3 on t1.uid = t3.uid
where t2.tag = 'SQL' and t1.score > 80
group by t3.level
order by level_cnt desc, level desc;