select
    user_profile.university as university,
    question_detail.difficult_level as difficult_level, 
    count(question_practice_detail.question_id) / count(distinct question_practice_detail.device_id) as avg_answer_cnt
from 
    question_practice_detail 
    INNER JOIN user_profile ON user_profile.device_id = question_practice_detail.device_id 
    INNER JOIN question_detail ON question_detail.question_id = question_practice_detail.question_id 
group by
    university,
    difficult_level

本题要点:

1.group by 分两步聚合

2.三表连接方法: 中间表

inner join 表1 on ____

inner join 表2 on ____