题解 | 试卷发布当天作答人数和平均分
select
er.exam_id,
count(distinct er.uid) uv,
round(avg(score), 1) avg_score
from
test.user_info ui
join test.exam_record er on ui.uid = er.uid
join test.examination_info ei on er.exam_id = ei.exam_id
where
tag = 'SQL'
and level > 5
group by
er.exam_id
order by
uv desc,
avg_score asc
