菜鸟一枚,莫名其妙懵对了 SELECT difficult_level,(sum(CASE WHEN result = 'right' THEN 1 ELSE 0 END ) / count(*)) corret_rate from user_profile p1 join question_practice_detail p2 JOIN question_detail p3 ON p1.device_id = p2.device_id AND p2.question_id = p3.question_id where university = '浙江大学' GROUP BY difficult_level ORDER BY corret_rate;