题解 | #统计每个学校各难度的用户平均刷题数#

select university,difficult_level,
(sum(cou_diff)/count(cou_diff)) as avg_answer_cnt
from(select university,difficult_level,cou_diff
from (select difficult_level,device_id,
count(difficult_level) as cou_diff
from (select device_id,a.question_id,difficult_level
from question_practice_detail as a 
inner join question_detail as b 
on a.question_id=b.question_id) as c 
group by difficult_level,device_id) as d
inner join user_profile as e 
on d.device_id=e.device_id ) as f
group by university,difficult_level
order by university,difficult_level