题解 | #统计每个学校各难度的用户平均刷题数#

select university,
        difficult_level,
        count(qpd.question_id)/count(distinct qpd.device_id) as avg_answer_cnt
        from question_practice_detail as qpd

         join user_profile  up on qpd.device_id=up.device_id


         join question_detail qd on qpd.question_id = qd.question_id


        group by university,difficult_level; 

题目很长，但描述得很详细，所以逻辑并不难，不要被吓倒。

count()计数，distinct去重，三表连接，与两表连接是类似的，只不过多加一句join...on...