题解 | #统计每个用户的平均刷题数#

①
select
    up.university,
    qd.difficult_level,
    round(
        count(qpd.question_id) / count(distinct qpd.device_id),
        4
    ) as avg_answer_cnt
from
    question_practice_detail as qpd
    inner join user_profile as up on up.device_id = qpd.device_id
    and up.university = '山东大学'
    inner join question_detail as qd on qd.question_id = qpd.question_id
group by
    qd.difficult_level;


②
SELECT
    up.university,
    qd.difficult_level,
    round(
        COUNT(qpd.device_id) / COUNT(DISTINCT qpd.device_id),
        4
    ) AS avg_answer_cnt
FROM
    user_profile up
    inner JOIN question_practice_detail qpd ON up.device_id = qpd.device_id
    inner JOIN question_detail qd ON qpd.question_id = qd.question_id
GROUP BY
    up.university,
    qd.difficult_level
HAVING
    up.university = '山东大学';