题解 | #异常的邮件概率#
异常的邮件概率
https://www.nowcoder.com/practice/d6dd656483b545159d3aa89b4c26004e
Select em.date, ROUND(COUNT(CASE WHEN em.type = 'no_completed' THEN 1 ELSE NULL END) / COUNT(*),3) AS completed_orders From email em left join user u1 on em.send_id = u1.id left join user u2 on em.receive_id = u2.id where u1.is_blacklist = 0 and u2.is_blacklist = 0 group by em.date order by em.date
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