题解 | #作答试卷得分大于过80的人的用户等级分布#

select `level`,count(uid) level_cnt
from(
	select ui.uid,`level`
	from examination_info ei inner join exam_record er 
		on ei.exam_id = er.exam_id 
		and tag = 'SQL'
		inner join user_info ui 
		on ui.uid = er.uid 
	group by ui.uid,`level` 
	having max(score) > 80
	) as 提取符合条件数据
group by `level`
order by level_cnt desc;