老规矩 打标签 sum统计 这类需求通杀

select 
    difficult_level,
     sum(if(result="right",1,0))/count(q.question_id) correct_rate
from question_practice_detail q,user_profile u,question_detail q2
where q.device_id=u.device_id 
	and q2.question_id=q.question_id 
	and university = "浙江大学"
group by difficult_level
order by correct_rate