题解 | #统计每个学校各难度的用户平均刷题数#

select
  e1.university,
 e3.difficult_level,
  count(e2.question_id)/count(distinct(e2.device_id)) as avg_answer_cnt
from
  user_profile as e1
  left join question_practice_detail as e2 on e1.device_id = e2.device_id
  left join question_detail as e3 on e2.question_id = e3.question_id
group by
  university,
 difficult_level having avg_answer_cnt !='None'
order by avg_answer_cnt
desc