题解 | #统计每个学校各难度的用户平均刷题数#

# 限定条件：无
# 每个学校：按照学校分组group by university
# 不同难度：按难度分组group by difficult_level
# 平均答题数：总答题数除以总人数count(qpd.question_id) / count(distinct qpd.device_id)
# 来自上面信息三个表，需要联表，up与qpd用device_id连接，qd与qpd用question_id连接。

#法一
select
    university,
    difficult_level,
    round(
        count(qpd.question_id) / count(distinct u.device_id),
        4
    ) as avg_answer_cnt
from
    user_profile as u,
    question_practice_detail as qpd,
    question_detail as qd
where
    u.device_id = qpd.device_id
    and qpd.question_id = qd.question_id
group by
    university,
    difficult_level

# 法二
select
    university,
    difficult_level,
    round(
        count(qpd.question_id) / count(distinct qpd.device_id),
        4
    ) as avg_answer_cnt
from
    question_practice_detail as qpd
    left join user_profile as up on up.device_id = qpd.device_id
    left join question_detail as qd on qd.question_id = qpd.question_id
group by
    university,
    difficult_level