# 查询每张sql试卷发布当天,五级以上的作答用户人数和平均分 select ei.exam_id, count(distinct er.uid) uv, round(avg(score),1) avg_score from examination_info ei left join exam_record er on datediff(er.submit_time, ei.release_time)<=1 and ei.exam_id=er.exam_id left join user_info u on er.uid=u.uid # 三表连接,同时表连接仅匹配了当天作答用户的答题情况 where ei.tag='SQL' # 筛选SQL类别的试卷,用户级别五级以上 and u.level>5 group by ei.exam_id order by uv desc, avg_score asc;