题解 | #统计每个学校各难度的用户平均刷题数#

SELECT up.university, qd.difficult_level, COUNT(qp.question_id) / COUNT(DISTINCT qp.device_id)
FROM question_practice_detail AS qp 
LEFT OUTER JOIN question_detail AS qd
ON qp.question_id = qd.question_id
LEFT OUTER JOIN user_profile AS up
ON qp.device_id = up.device_id 
GROUP BY up.university, qd.difficult_level;

联结三个表，以及按两个字段分组+聚合