题解 | #统计每个学校各难度的用户平均刷题数#

SELECT u.university, qd.difficult_level, COUNT(p.question_id) /count(distinct p.device_id) as avg_answer_cnt FROM question_practice_detail p JOIN question_detail qd on qd.question_id = p.question_id JOIN user_profile u on u.device_id = p.device_id GROUP BY university,difficult_level