select user_profile.university as university, question_detail.difficult_level as difficult_level, count(question_practice_detail.question_id) / count(distinct question_practice_detail.device_id) as avg_answer_cnt from question_practice_detail INNER JOIN user_profile ON user_profile.device_id = question_practice_detail.device_id INNER JOIN question_detail ON question_detail.question_id = question_practice_detail.question_id group by university, difficult_level
本题要点:
1.group by 分两步聚合
2.三表连接方法: 中间表
inner join 表1 on ____
inner join 表2 on ____