清爽的解法 | #试卷发布当天作答人数和平均分#

SELECT
    exam_id,
    COUNT(DISTINCT uid) as uv,
    ROUND(SUM(score)/COUNT(DISTINCT r.id),1) as avg_score
FROM
    user_info
JOIN
    exam_record r USING(uid)
JOIN
    examination_info USING(exam_id)
WHERE
    tag = 'SQL' and level > 5 and DAY(release_time) = DAY(submit_time)
GROUP BY
    exam_id
ORDER BY
    uv DESC, avg_score