题解 | #试卷发布当天作答人数和平均分#

select
    exam_id,
    count(distinct uid) uv,
    round(avg(score), 1) avg_score
from
    exam_record er
    left join user_info ui using (uid)
    left join examination_info ei using (exam_id)
where
    tag = 'SQL'
    and date_format (start_time, '%Y%m%d') = date_format (release_time, '%Y%m%d')
    and level > 5
group by
    exam_id
order by
    uv desc,
    avg_score asc;