题解 | #异常的邮件概率#

select date,round(1-sum(if(type='completed',1,0))/count(*),3)
from email e left join user u1 on e.send_id=u1.id 
left join user u2 on e.receive_id=u2.id
where u1.is_blacklist='0' and u2.is_blacklist='0'
group by date
order by date