题解 | #试卷发布当天作答人数和平均分#

select exam_id, count(distinct uid) uv,round(avg(score),1) avg_score
from
(
select a.uid,b.exam_id,score
from user_info a,examination_info b,exam_record c 
where a.uid = c.uid
and b.exam_id = c.exam_id
and date(c.start_time) = date(b.release_time)
and tag = 'SQL'
and level > 5
and submit_time is not null
) d
group by exam_id
order by uv desc,avg_score;