基础练习
select t3.university ,t4.difficult_level,
count(t3.question_id)/count(distinct t3.device_id) as avg_answer_cnt
from (
select t1.university,t2.question_id,t1.device_id
from user_profile t1 right join question_practice_detail t2
on t1.device_id=t2.device_id
where t1.university='山东大学'
)t3
inner join question_detail t4
on t3.question_id=t4.question_id
group by t3.university,t4.difficult_level;
先是内连接获取我想要的数据，变成一个表继续内连接，即子查询
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