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(count(qpd.id)/count(distinct up.device_id),4) as avg_answer_cnt
from user_profile up
left join question_practice_detail qpd on up.device_id=qpd.device_id
left join question_detail qd on qpd.question_id=qd.question_id
group by up.university, qd.difficult_level
having qd.difficult_level is not null;--只保留有实际难度的统计结果