题解 | #统计每个用户的平均刷题数#

SELECT '山东大学' AS university, difficult_level, ROUND(COUNT(qpd.question_id)/COUNT(DISTINCT qpd.device_id),4) AS avg_answer_cnt
FROM question_practice_detail AS qpd

INNER JOIN user_profile AS up
ON up.device_id = qpd.device_id AND university = '山东大学'

INNER JOIN question_detail AS qd
ON qd.question_id = qpd.question_id

GROUP BY difficult_level