题解 | #异常的邮件概率#
异常的邮件概率
https://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select rk.date, round(sum(case when type = 'no_completed' then 1 else 0 end)/count(type),3) p from (select e.id, e.send_id, e.receive_id, e.type, e.date, u.is_blacklist send_posn from email e join user u on e.send_id = u.id) rk join (select e.id, u.is_blacklist receive_posn from email e join user u on e.receive_id = u.id) ak on rk.id = ak.id where rk.send_posn != 1 and ak.receive_posn != 1 group by rk.date order by rk.date
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