题解 | #统计每个用户的平均刷题数#

先把山东大学的id查出来，然后再去找其他的

select 
   university,
   difficult_level,
   count(t2.question_id)/count(distinct t2.device_id)
FROM 
    (SELECT university,device_id FROM user_profile WHERE university='山东大学' ORDER by university) as t1,
    question_practice_detail as t2,
    question_detail as qd
where 
    t1.device_id = t2.device_id and t2.question_id = qd.question_id
group by 
    qd.difficult_level	;