题解 | 统计每个用户的平均刷题数

select

up.university,

qd.difficult_level,

round(count(qpd.question_id) / count(distinct(qpd.device_id)),4) 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 university,difficult_level

having up.university = "山东大学"