题解 | #异常的邮件概率#
异常的邮件概率
https://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select e.date,round(if(c2 is null,0,c2)/c1,3) p from (select date,count(type) c1 from email where send_id in(select id from user where is_blacklist=0) and receive_id in(select id from user where is_blacklist=0) group by date) e left join (select date,count(type) c2 from email where type='no_completed' and send_id in(select id from user where is_blacklist=0) and receive_id in(select id from user where is_blacklist=0) group by date ) en on e.date=en.date order by date
