题解 | 统计每个学校的答过题的用户的平均答题数
SELECT university, COUNT(q.question_id)/COUNT(DISTINCT q.device_id) avg_answer_cnt FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id GROUP BY university ORDER BY university
逻辑上有点绕
SELECT university, COUNT(q.question_id)/COUNT(DISTINCT q.device_id) avg_answer_cnt FROM question_practice_detail q LEFT JOIN user_profile u ON q.device_id = u.device_id GROUP BY university ORDER BY university
逻辑上有点绕
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