题解 | 统计每个学校各难度的用户平均刷题数
SELECT university,
q2.difficult_level,
ROUND(COUNT(q1.question_id)/COUNT(DISTINCT q1.device_id), 4) AS avg_answer_cnt
FROM user_profile u
INNER JOIN question_practice_detail q1 ON u.device_id = q1.device_id
INNER JOIN question_detail q2 ON q1.question_id = q2.question_id
GROUP BY university, q2.difficult_level
ORDER BY university, q2.difficult_level
记得看下表3里面的difficult_level打错了
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