题解 | #试卷发布当天作答人数和平均分#

select
    exam_record.exam_id,
    count(distinct exam_record.uid) as uv,
    format (avg(score), 1) as avg_score
from
    (
        select
            exam_id,
            date (release_time) as date
        from
            examination_info
        where
            tag = 'SQL'
    ) as x_sql
    join exam_record on x_sql.exam_id = exam_record.exam_id
    join user_info on exam_record.uid = user_info.uid
where
    date (start_time) = x_sql.date
    and level > 5
group by
    exam_record.exam_id
order by
    uv desc,
    avg_score;