select
qd.difficult_level,
sum(if(result = 'right',1,0))/count(result) as correct_rate
from
question_practice_detail as upd
left join user_profile as up on up.device_id = upd.device_id
left join question_detail as qd on qd.question_id = upd.question_id
where
university = '浙江大学'
group by
qd.difficult_level
order by correct_rate asc



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