题解 | #统计每个学校各难度的用户平均刷题数#

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