select qd.difficult_level, round(AVG(if(q.result = 'right',1,0)),4) as correct_rate from user_profile u join question_practice_detail q on u.device_id = q.device_id join question_detail qd on q.question_id = qd.question_id where u.university = '浙江大学' group by qd.difficult_level order by correct_rate ASC;
这里使用left join左连接会多出现一排为None的结果。