题解 | #统计每个用户的平均刷题数#

select a.university,c.difficult_level,
(count(b.question_id)/count(distinct(a.device_id))) as avg_answer_cnt
from user_profile as a,
question_practice_detail as b,
question_detail as c
where a.device_id=b.device_id 
and b.question_id=c.question_id
group by a.university,c.difficult_level
having university='山东大学'