select qd.difficult_level,
round(AVG(if(q.result = 'right',1,0)),4) as correct_rate
from
user_profile u join question_practice_detail q
on u.device_id = q.device_id
join question_detail qd
on q.question_id = qd.question_id
where u.university = '浙江大学'
group by
qd.difficult_level
order by
correct_rate
ASC;
这里使用left join左连接会多出现一排为None的结果。



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