题解 | #试卷发布当天作答人数和平均分#

select exam_record.exam_id,count(distinct exam_record.uid),round(avg(score),1)
from exam_record 
left join examination_info on exam_record.exam_id=
examination_info.exam_id
left join user_info on user_info.uid=exam_record.uid
where level>5
and date_format(start_time,'%Y-%m-%d')=date_format(release_time,'%Y-%m-%d')
group by exam_record.exam_id
order by count(distinct exam_record.uid) desc,
avg(score)