sql查询优先级
1.from
2.on
3.join
4.where
5.group by
6.having
7.select !!!!!!!!!
8.distinct
9.order by 
SELECT difficult_level, 
    sum(if(result = 'right',1,0)) / count(result) as correct_rate
FROM user_profile t1
INNER join question_practice_detail t2
on t1.device_id = t2.device_id 
INNER JOIN question_detail t3
on t2.question_id = t3.question_id
where university = '浙江大学'
group by difficult_level
order by correct_rate