题解 | #异常的邮件概率#
异常的邮件概率
http://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
Mysql的写法：
select date,
round((
select count() from
(
select date,type
from email e
where e.send_id in
(
select id from user
where user.is_blacklist=0
)
and e.receive_id in
(
select id from user
where user.is_blacklist=0
)
)as n_u1
where n_u1.date=normal_user.date and n_u1.type='no_completed'
)1.0/count(type),3 )
from
(
select date,type
from email e
where e.send_id in
(
select id from user
where user.is_blacklist=0
)
and e.receive_id in
(
select id from user
where user.is_blacklist=0
)
)as normal_user
group by date order by date asc
