题解 | #统计每个用户的平均刷题数#

select
    t1.university,t3.difficult_level,
    COUNT(t2.question_id) / COUNT(DISTINCT(t2.device_id)) as avg_answer_cnt
from
    user_profile as t1,
    question_practice_detail as t2,
    question_detail as t3
where
    t1.university = '山东大学'
    and t1.device_id = t2.device_id
    and t2.question_id = t3.question_id
group by
    t3.difficult_level;