题解 | #统计每个学校各难度的用户平均刷题数#

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

• 每个学校：按学校分组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。