题解 | #异常的邮件概率#
异常的邮件概率
http://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select
date,
round(count(if(type='no_completed',true,null))/count(em.id) ,3) as p
from email em
left join user u1 on em.send_id=u1.id
left join user u2 on em.receive_id=u2.id
where u1.is_blacklist=0 and u2.is_blacklist=0
group by date
order by date asc
查看8道真题和解析