select 
    difficult_level,
    round(
        sum(
            case 
                when qpd.result = "right" then 1
                else 0
                end
        )/count(qpd.question_id),4
        )
    as correct_rate
from user_profile up#记为up表
join question_practice_detail qpd#记为qpd表
on up.device_id = qpd.device_id
join question_detail qd #记为qd表
on qpd.question_id = qd.question_id
where university = "浙江大学" 
group by qd.difficult_level
order by correct_rate asc