题解 | #统计每个学校各难度的用户平均刷题数#

select
    university,
    difficult_level,
	#注意distinct qpd.device_id
    round(
        count(qpd.question_id) / count(distinct qpd.device_id),
        4
    ) as avg_answer_cnt
from
    user_profile up,
    question_practice_detail qpd,
    question_detail qd
where
    up.device_id = qpd.device_id
    and qpd.question_id = qd.question_id
group by
    university,
    difficult_level;