题解 | #试卷发布当天作答人数和平均分#

select er.exam_id
        ,count(distinct er.uid) AS uv
        ,round(avg(er.score),1) AS avg_score
from examination_info ei
left join exam_record er on ei.exam_id = er.exam_id
left join user_info ui on er.uid = ui.uid 
where tag = 'SQL' and level>5 and date(ei.release_time) = date(er.submit_time)
group by er.exam_id
order by count(distinct er.uid) desc
        ,round(avg(er.score),1) asc