题解 | 统计每个学校各难度的用户平均刷题数

select
  university,difficult_level,round(count(qpd.question_id)/count(distinct qpd.device_id),4) as avg_answer_cnt
from
    user_profile as up inner join question_practice_detail as qpd on up.device_id = qpd.device_id 
    inner join question_detail as qd on qpd.question_id = qd.question_id  
group by university,difficult_level
order by university,difficult_level

计算平均值可用一般用count提取个数，列较为简单时使用avg函数

join可以连续使用 如本题连接三个表