题解 | #异常的邮件概率#

select e.date,round(sum(case e.type when 'no_completed' then 1 else 0 end)
                    /count(*),3)
from email e 
join user u1 on e.receive_id = u1.id and u1.is_blacklist <> 1
join user u2 on e.send_id = u2.id and u2.is_blacklist <> 1
group by e.date
order by e.date