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(b.device_id))) as avg_answer_cnt
from
    user_profile a,
    question_practice_detail b,
    question_detail c
where
    a.university = '山东大学'
    AND a.device_id = b.device_id
    AND b.question_id = c.question_id
GROUP BY difficult_level