select
qd.difficult_level,
round(
sum(
case
when qpd.result = 'right' then 1
else 0
end
) / count(qpd.question_id),
4
) as correct_rate
from
user_profile up
join
question_practice_detail qpd
on up.device_id = qpd.device_id
join
question_detail qd
on qd.question_id = qpd.question_id
where
up.university in ('浙江大学')
group by
qd.difficult_level
order by
correct_rate
三个表,先取出浙江大学的数据,再关联答题记录,最后关联题目难度



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