就这?就这?3分钟不能再多了
select
difficult_level,
count(if(p.result = 'right',1,NULL)) / count(p.result) correct_rate
FROM
question_practice_detail p
JOIN(
SELECT
university,device_id
from user_profile
WHERE university= '浙江大学'
) u
ON u.device_id = p.device_id
JOIN question_detail d
ON d.question_id = p.question_id
GROUP BY
difficult_level
ORDER BY
correct_rate