题解 | #作答试卷得分大于过80的人的用户等级分布#

select d.level,count(d.score)level_cnt

from

(select a.uid,a.exam_id,a.score,

DATE_FORMAT(a.start_time,'%Y-%m-%d')start_time,

DATE_FORMAT(a.submit_time,'%Y-%m-%d')submit_time,

b.tag,c.level

from

exam_record a

join 

examination_info b

on a.exam_id=b.exam_id

join

user_info c

on a.uid=c.uid)d

where d.tag='SQL' and d.score>80

group by d.level

order by level_cnt desc;