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 u.device_id),4) as avg_answer_cnt
from user_profile as u

inner join question_practice_detail as q 
on q.device_id = u.device_id

inner join question_detail as qd
on q.question_id = qd.question_id

group by university,difficult_level

一开始就想的是用inner join,但是苦于不知道怎么写，以为需要重新选择交集表再写一次，看了答案才知道，可以连着写inner join,不需要再选择一次对应表列