题解 | #浙大不同难度题目的正确率#

SELECT
	t3.difficult_level,
	SUM( result = 'right' ) / COUNT( t1.device_id ) AS correct_rate 
FROM
	question_practice_detail AS t1
	LEFT JOIN user_profile AS t2 ON t1.device_id = t2.device_id
	LEFT JOIN question_detail AS t3 ON t1.question_id = t3.question_id 
WHERE
	t2.university = '浙江大学' 
GROUP BY
	t3.difficult_level
ORDER BY correct_rate ASC

我感觉这种直接明了，没那么绕