题解 | #统计每个用户的平均刷题数#

select
u.university,
qd.difficult_level,
count(q.result)/count(distinct(q.device_id))as avg_question_cnt
from user_profile as u
left join question_practice_detail as q
on u.device_id=q.device_id
left join question_detail as qd
on q.question_id=qd.question_id
group by 
u.university,
qd.difficult_level
having u.university ='山东大学'