题解 | #统计每个学校各难度的用户平均刷题数#

SELECT  university, 
        difficult_level,
        ROUND(COUNT(*) / COUNT(DISTINCT u.device_id), 4) AS avg_answer_cnt
FROM    question_practice_detail p, 
        user_profile u,
        question_detail d
WHERE   u.device_id = p.device_id 
        AND p.question_id = d.question_id
GROUP BY    university, difficult_level