题解 | #统计每个学校各难度的用户平均刷题数#

方法1：三表左外连接

select
    c.university,
    b.difficult_level,
    count(a.device_id) / count(distinct a.device_id) as avg_answer_cnt
from
    question_practice_detail a
    left join question_detail b on a.question_id = b.question_id
    left join user_profile c on a.device_id = c.device_id
group by
    c.university,
    b.difficult_level

方法2：三表内连接

select
    c.university,
    b.difficult_level,
    count(a.device_id) / count(distinct a.device_id) as avg_answer_cnt
from
    question_practice_detail a,
    question_detail b,
    user_profile c
where a.question_id = b.question_id and a.device_id = c.device_id
group by
    c.university,
    b.difficult_level