题解 | #统计每个用户的平均刷题数#

select 
    u.university,
    qd.difficult_level,
    count(q.question_id) / count(distinct q.device_id) as avg_answer_cnt
from 
    user_profile as u,
    question_practice_detail as q,
    question_detail as qd
where 
    u.university = "山东大***表1 表2
    and u.device_id = q.device_id
    # 联表2 表3
    and q.question_id = qd.question_id
    
group by 
    university, difficult_level