select difficult_level, sum(result='right')/count(*) as correct_rate from
(select user_profile.device_id, question_id, result, university from user_profile
inner join
question_practice_detail
on user_profile.device_id=question_practice_detail.device_id where university = '浙江大学') a
inner join
question_detail
on
a.question_id = question_detail.question_id
group by difficult_level
order by correct_rate;