select
    qd.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
join 
    question_practice_detail qpd 
    on up.device_id = qpd.device_id
join 
    question_detail qd 
    on qd.question_id = qpd.question_id

where
    up.university in ('浙江大学')
group by
    qd.difficult_level
order by
    correct_rate

三个表,先取出浙江大学的数据,再关联答题记录,最后关联题目难度