# 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;