题解 | #试卷发布当天作答人数和平均分#

select exam_id,count(distinct uid) as uv,round(sum(score)/count(score),1) as avg_score from exam_record join examination_info using(exam_id) join user_info using(uid) where tag='SQL' and date(release_time)=date(start_time) and level>5 group by exam_id order by uv desc,avg_score