题解 | #统计每个用户的平均刷题数#

select 
    u.university,q2.difficult_level,(count(q1.question_id)/count(distinct(q1.device_id))) as avg_answer
    from
    user_profile as u
    inner join
    question_practice_detail as q1
    on
    u.device_id = q1.device_id
    inner join
    question_detail as q2
    on
    q1.question_id=q2.question_id
    group by u.university, q2.difficult_level
    having u.university = "山东大学"