题解 | #统计每个学校各难度的用户平均刷题数#

select u.university, qd.difficult_level, count(qd.question_id) /count(distinct u.device_id) as avg_answer_cnt
from user_profile u, question_practice_detail q, question_detail qd
where u.device_id = q.device_id and q.question_id = qd.question_id
group by qd.difficult_level, u.university
order by u.university;