大神们帮忙看看这样合理不? select a.university, c.difficult_level, round(count(b.question_id) / count(distinct b.device_id),4 ) as avg_answer_cnt from question_practice_detail as b left join user_profile a on b.device_id = a.device_id left join question_detail as c on b.question_id = c.question_id where a.university='山东大学' group by c.difficult_level
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Cherrycola01:0实习 0项目 约等于啥也没有啊 哥们儿这简历认真的吗
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