题解 | #统计每个用户的平均刷题数#

SELECT 
    university,
    difficult_level,
    ROUND(COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id),4) AS avg_answer_cnt
FROM question_practice_detail qpd
LEFT JOIN user_profile up
    USING(device_id)
    
LEFT JOIN question_detail qd
    USING(question_id)

GROUP BY university,difficult_level
HAVING university="山东大学";