题解 | #试卷发布当天作答人数和平均分#

select 
    t2.exam_id,
    count(distinct t3.uid) as uv,
    round(avg(t3.score),1) as avg_score
from
    user_info t1,
    examination_info t2,
    exam_record t3
where
    t3.uid = t1.uid
    and t3.exam_id = t2.exam_id
    and t2.tag = 'SQL'
    and t3.start_time >= t2.release_time
    and t3.score is not null
    and t1.level > 5
group by
    t2.exam_id
order by
    uv desc,avg_score