题解 | 统计每个用户的平均刷题数

select p.university,m.difficult_level,
    round(count(m.question_id)/count(distinct(p.device_id)),4) as avg_answer_cnt
from user_profile p
join question_practice_detail t
    on p.device_id=t.device_id
join question_detail m
    on t.question_id = m.question_id
where university = "山东大学"
group by university,difficult_level

where university = "山东大学"

多了个分组前筛选