题解 | #作答试卷得分大于过80的人的用户等级分布#

SELECT `level`, COUNT(*) level_cnt
FROM user_info
WHERE uid IN(
	SELECT uid
	FROM 
		exam_record
	WHERE 
		exam_id IN (
			SELECT exam_id FROM examination_info WHERE tag = 'SQL'
		) AND score > 80
) GROUP BY `level`
ORDER BY level_cnt DESC;