select bb.university, cc.difficult_level, count(aa.device_id)/count (distinct aa.device_id) as avg_answer_cnt from question_practice_detail aa left join user_profile bb on aa.device_id=bb.device_id left join question_detail cc on aa.question_id=cc.question_id group by bb.university, cc.difficult_level having bb.university='山东大学' 我这个凭什么报错
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