题解 | #浙大不同难度题目的正确率#

select difficult_level, 
sum(case when result='right' then 1 else 0 end)/sum(case when difficult_level='easy' then 1 when difficult_level='medium' then 1 when difficult_level='hard' then 1 else 0 end) as correct_rate
from question_practice_detail as qpd inner join question_detail as qd on qpd.question_id = qd.question_id
where device_id in(
    select device_id from user_profile where university = '浙江大学')
group by difficult_level
order by correct_rate asc

感觉还挺直来直去的