问题分解: 1.浙江大学的用户 university='浙江大学' 2.不同难度 group by difficult_level 3.正确率顺序排序 order by correct_rate 4.正确率 sum(if(result='right',1,0) / count(question_id) as correct_rate

select
    difficult_level,sum(if(result='right',1,0)) / count(d.question_id) as correct_rate
from
    question_practice_detail p
join
    user_profile u
ON
    p.device_id = u.device_id
join
    question_detail d
on
    p.question_id = d.question_id
where
    university='浙江大学'
group by 
    difficult_level
order by
    correct_rate