SELECT
qd.difficult_level,
SUM(IF (result = 'right', 1, 0)) / COUNT(qpd.device_id) AS correct_rate
FROM
question_detail AS qd
LEFT JOIN question_practice_detail AS qpd ON qpd.question_id = qd.question_id
LEFT JOIN user_profile AS up ON up.device_id = qpd.device_id
WHERE
up.university = '浙江大学'
GrOUP BY
qd.difficult_level
ORDER BY
correct_rate ASC;



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