题解 | #统计每个学校各难度的用户平均刷题数#

等值连接

    university,
    difficult_level,
    count(qpd.question_id)/count(distinct qpd.device_id) as avg_answer
from 
question_practice_detail as qpd
,user_profile as up 
,question_detail as qd
where qd.question_id = qpd.question_id
and up.device_id = qpd.device_id

group by university, difficult_level