题解 | #浙大不同难度题目的正确率#

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

一个比较易于理解的解法