题解 | #统计每个用户的平均刷题数#

select a.university,c.difficult_level,count(b.question_id) / count(distinct(a.device_id)) as avg_answer_cnt FROM question_practice_detail b left join user_profile a on b.device_id=a.device_id left join question_detail c on b.question_id=c.question_id where a.university='山东大学' and b.result is not null group by a.university,c.difficult_level