题解 | #作答试卷得分大于过80的人的用户等级分布#

SELECT
    level,
    COUNT(IF(score>80,uid,null)) AS level_cnt
FROM exam_record 
JOIN examination_info USING(exam_id)
JOIN user_info USING(uid)
WHERE tag = 'SQL'
GROUP BY level
HAVING level_cnt > 0
ORDER BY level_cnt DESC;