题解 | #试卷发布当天作答人数和平均分#

select exam_id,count(distinct exam_record.uid) as uv,round(avg(score),1) as avg_score from user_info join exam_record on user_info.uid=exam_record.uid where submit_time is not null and exam_id in (select exam_id from examination_info where tag='SQL') and date_format(submit_time,'%Y%m%d') in (select date_format(release_time,'%Y%m%d') from examination_info where tag='SQL') and level>5 group by exam_id order by uv desc, avg_score;