题解 | #统计每个学校各难度的用户平均刷题数#

SELECT university,difficult_level, COUNT(qp.question_id) / COUNT(DISTINCT u.device_id) AS avg_answer_cnt

FROM user_profile u, question_practice_detail qp, question_detail qd

WHERE u.device_id = qp.device_id

AND qp.question_id = qd.question_id

AND question_cnt > 0

GROUP BY university, difficult_level

直接表连接就行啦，用LEFT JOIN 感觉多此一举了