题解 | 统计每个学校各难度的用户平均刷题数

select
    u.university,
    qd.difficult_level,
    round(count(qpd.id) / count(distinct u.device_id), 4) as avg_answer_cnt
from
    user_profile u
    join question_practice_detail qpd on u.device_id = qpd.device_id
    join question_detail qd on qpd.question_id = qd.question_id
group by
    u.university,
    qd.difficult_level
order by
    u.university,
    qd.difficult_level;

SELECT
    u.university,
    qd.difficult_level,
    ROUND(COUNT(qpd.id) / COUNT(DISTINCT u.device_id), 4) AS avg_answer_cnt
FROM
    user_profile u
    JOIN question_practice_detail qpd ON u.device_id = qpd.device_id
    JOIN question_detail qd ON qpd.question_id = qd.question_id
GROUP BY
    u.university,
    qd.difficult_level
ORDER BY
    u.university,
    qd.difficult_level;

-- 要了解所查询的字段是从哪里来的，怎么来的，要用到什么函数
-- 表之间的关联是靠什么
-- 根据什么来分组排序