题解 | #统计每个用户的平均刷题数#

select
    user_profile.university as university,
    question_detail.difficult_level as difficult_level,
    round(count(question_practice_detail.question_id)/count(distinct question_practice_detail.device_id),4) as avg_answer_cnt
from 
    user_profile,question_practice_detail,question_detail
where 
    user_profile.device_id=question_practice_detail.device_id and
    question_practice_detail.question_id=question_detail.question_id and
    university='山东大学'
group by
    university,difficult_level;