题解 | #统计每个用户的平均刷题数#

读懂题目的隐含意，分组包含了学校分组和难度分组，最后使用having分组

select 
    university,
    difficult_level,
    count(qpd.question_id) / count(distinct qpd.device_id) 
    as avg_answer_cnt
from question_practice_detail as qpd

left join user_profile as up
on up.device_id=qpd.device_id

left join question_detail as qd
on qd.question_id=qpd.question_id

group by difficult_level, university
having university = '山东大学'