SELECT qd.difficult_level,
# 这里分子用 sum,求该列上所有值的和;分母用 count,求有多少个非空行
sum(if(qpd.result = 'right', 1, 0)) / count(qpd.question_id) correct_rate
# AVG(IF(result='right',1,0)) AS correct_rate
from user_profile u
join question_practice_detail qpd
on u.device_id = qpd.device_id
join question_detail qd
on qpd.question_id = qd.question_id
where u.university = '浙江大学'
group by difficult_level
order by correct_rate;