题解 | #试卷发布当天作答人数和平均分#

    select
        i.exam_id
        ,r.uid
        ,r.score
    from examination_info i join exam_record r using(exam_id)
    where i.tag='SQL' and substr(i.release_time,1,10)=substr(r.start_time,1,10)
)
select 
    t1.exam_id
    ,count(distinct t1.uid) as uv
    ,round(avg(t1.score),1) as avg_score
from user_info u join t1 using(uid)
where u.level>5
group by t1.exam_id
order by uv DESC
        ,avg_score