题解 | 统计每个用户的平均刷题数

select
    university,
    difficult_level,
    count(qpd.question_id)/count(distinct qpd.device_id) as avg_answer_cnt
from question_practice_detail as qpd
inner join question_detail as qd
on qd.question_id = qpd.question_id
inner join user_profile as up
on up.device_id = qpd.device_id
group by university,difficult_level
having up.university = '山东大学'

其实没太懂，是只出现一次的label就不需要加表格名.限制吗？