# 五级以上用户作答SQL试卷的信息 SELECT ei.exam_id, COUNT(DISTINCT er.uid) AS uv, ROUND(AVG(er.score), 1) AS avg_score # 计算人数和平均分 FROM exam_record AS er LEFT OUTER JOIN examination_info AS ei ON er.exam_id = ei.exam_id LEFT OUTER JOIN user_info AS u ON er.uid = u.uid WHERE ei.tag = 'SQL' AND u.level > 5 AND DA...