题解 | #统计每个用户的平均刷题数#

select university, difficult_level, count(qd.question_id)/count(DISTINCT qpd.device_id) as avg_answer_cnt from user_profile as u inner join question_practice_detail as qpd on u.device_id = qpd.device_id inner join
question_detail as qd on qd.question_id = qpd.question_id where university = '山东大学' group by difficult_level;