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(avg(score),1)as avg_score
from user_info join
(select exam_id,score,uid
from exam_record
where exam_id in(
select exam_id
from examination_info
where tag ='SQL')) as erc
using(uid)
where user_info.level>5
group by exam_id
order by uv desc,avg_score