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 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

1. 不同学校不同难度 group by university, difficult_level
2. 平均答题数：总答题数除以总人数 round(count(qpd.question_id) / count(distinct qpd.device_id), 4) as avg_answer_cnt
3. 平均值精度：保留4位小数round(x, 4)
4. 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