题解 | #月均完成试卷数不小于3的用户爱作答的类别#

select tag,count(*) tag_cnt 
from exam_record a 
left join examination_info b 
on a.exam_id=b.exam_id 
where uid 
in (
select uid from
(
select uid, any_value(cnt)/count(month) avg
from
(
select uid,count(tag) cnt,month
from 
(
select uid,month(start_time) month,tag 
from exam_record a 
left join examination_info b 
on a.exam_id=b.exam_id 
where submit_time is not null
)c 
group by uid,month
) d
group by uid,month
) e
where avg >=3
) 
group by tag
order by count(tag) desc;