题解 | 统计每个学校各难度的用户平均刷题数

SELECT
user_profile.university,
question_detail.difficult_level,
round(count(question_practice_detail.question_id)/count(DISTINCT question_practice_detail.device_id),4)as avg_answer_cnt
from user_profile 
inner join question_practice_detail
on user_profile.device_id = question_practice_detail.device_id
inner join question_detail
on question_practice_detail.question_id = question_detail.question_id
group by university,difficult_level;

inner join = join

好多人写left join ? 那岂不是保留左边表 ，未答题的用户也会被连接

但是在计算是会添加question_practice_detail.（不加有歧义），所以对结果应该是没有影响的，因为第二个表一定是答题用户