题解 | #统计每个用户的平均刷题数#

select university,
       difficult_level,
       round(count(t2.device_id)/count(distinct t2.device_id),4) as avg_answer_cnt
from user_profile t1,
    (select q1.device_id,q1.question_id,q2.difficult_level
        from question_practice_detail q1,question_detail q2
        where q1.question_id=q2.question_id) as t2
where t1.device_id = t2.device_id
group by university,difficult_level
having university="山东大学"

和SQL23一样 groupy by 后 having一个山东大学的条件就可以的