题解 | #作答试卷得分大于过80的人的用户等级分布#

SELECT ui.level,COUNT(ui.uid) AS level_cnt FROM user_info ui JOIN exam_record er ON ui.uid=er.uid WHERE er.exam_id IN (SELECT exam_id FROM examination_info WHERE tag='SQL') AND er.score> 80 GROUP BY ui.level ORDER BY level_cnt DESC;

题目很简单，不难，慢慢做，捋清楚。