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 er.exam_id,er.uid,score
from examination_info ei inner join exam_record er
	on ei.exam_id = er.exam_id 
	and tag = 'SQL'
	and left(release_time,10) = left(start_time,10)
	inner join user_info ui 
	on ui.uid = er.uid
	and `level` > 5	
) as 提取符合条件的数据
group by exam_id
order by uv desc,avg_score;