#沿用上一题热评第一大佬的思路#
select qd.difficult_level,
       sum(if(qpd.result='right',1,0))/sum(if(qpd.result='right',1,1)) as correct_rate
from user_profile up
join question_practice_detail qpd
using(device_id)
join question_detail qd
on qpd.question_id=qd.question_id
where up.university = '浙江大学'
group by qd.difficult_level
order by correct_rate;