题解 | #试卷发布当天作答人数和平均分#

select
  t1.exam_id,
  count(distinct t3.uid) as uv,
  round(sum(t1.score) / count(t1.exam_id), 1) as avg_score
from
  exam_record t1
  left join examination_info t2 on t1.exam_id = t2.exam_id
  left join user_info t3 on t1.uid = t3.uid
where
  t1.submit_time is not null
  and t2.tag = 'SQL'
  and t3.level > 5
group by
  t1.exam_id
order by
  uv desc,
  avg_score asc