题解 | #统计每个学校各难度的用户平均刷题数#

select 
    university,
    difficult_level,
    round(count(q.question_id)/count(distinct q.device_id),4) as avg_answer_cnt
from user_profile as u 
right join question_practice_detail as q 
on u.device_id=q.device_id
right join question_detail as d
on q.question_id=d.question_id
group by u.university,d.difficult_level;

select

    university,

    difficult_level,

   round(count(q.question_id)/count(distinct q.device_id),4)

from question_practice_detail as q

left join user_profile as u

on u.device_id=q.device_id

right join question_detail as d

on q.question_id=d.question_id

group by u.university,d.difficult_level;