题解 | #异常的邮件概率#
异常的邮件概率
http://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select e.date,
round(sum(case when e.type='completed' then 0 else 1 end)/count(e.id),3) as p
from email as e inner join user as u1 on e.send_id=u1.id inner join user as u2 on e.receive_id=u2.id
where u1.is_blacklist<>1 and u2.is_blacklist<>1
group by e.date
order by e.date asc
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