题解 | #统计每个学校各难度的用户平均刷题数#

select
    up.university,
    qd.difficult_level,
	#round函数四舍五入取后4为小数,
    round(count(qpd.question_id) / count(distinct qpd.device_id),4) as avg_answer_cnt
from 
    user_profile as up,
    question_practice_detail as qpd,
    question_detail as qd
where
    up.device_id = qpd.device_id and qpd.question_id = qd.question_id
group by 
    up.university,qd.difficult_level;