题解 | 浙大不同难度题目的正确率

select difficult_level,
        # count(question_id) as all_question,
        sum(case when result='right' then 1 else 0 end)/count(question_id) as correct_rate
from (
select qd.difficult_level,
        pd.question_id,
        pd.result
from user_profile up
left join question_practice_detail pd on pd.device_id=up.device_id
left join question_detail qd on qd.question_id=pd.question_id
where up.university='浙江大学'
) t
where difficult_level is not null
group by difficult_level
order by correct_rate asc

还得把difficult_level为空的给过滤掉，老方法了，在项目导数的时候，如果说有重复的数据，我就直接加distinct去重，有啥问题解决问题，以问题为导向型编程