题解 | #统计每个用户的平均刷题数#

select university,
difficult_level,
round(count(qpd.question_id)/count(distinct u.device_id),4) 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 university,difficult_level having university='山东大学';
涉及到我对于group by的一个理解：将相同的值进行一个汇总。为什么不用left join 因为会把空值也包含进去了。