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(qpd.question_id) / COUNT(DISTINCT qpd.device_id),4) AS avg_answer_cnt
FROM question_practice_detail  qpd

LEFT JOIN user_profile  up 
ON up.device_id=qpd.device_id

LEFT JOIN question_detail  qd 
ON qd.question_id=qpd.question_id

GROUP BY university,difficult_level;
//LEFT JOIN 进行连接