知识点：left join
select 
difficult_level,
avg(if(result='right',1,0)) as correct_rate
from (
select device_id
from user_profile 
where university='浙江大学'
)a 
left join question_practice_detail b
on a.device_id=b.device_id 
left join question_detail c
on b.question_id=c.question_id
where difficult_level is not null
group by difficult_level
order by correct_rate
;