题解 | #试卷发布当天作答人数和平均分#

select
a.exam_id,
count(distinct a.uid) as uv,
round(avg(a.score),1) as avg_score
from exam_record a
join examination_info b
on a.exam_id = b.exam_id
where b.tag ='SQL'
AND a.uid in (select uid from user_info where level >5)
and date(a.start_time)=date(b.release_time)
group by a.exam_id
order by uv desc,avg_score