题解 | 异常的邮件概率
异常的邮件概率
https://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
select date, round(count(if(type='no_completed',type,null))/count(type),3) p from email e join user u1 on e.send_id = u1.id 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
count嵌套if函数，满足条件统计，否则null代表不统计
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