题解 | #异常的邮件概率#

SELECT e.date

    ,ROUND(SUM(IF(e.type="no_completed",1,0))/COUNT(*),3)

FROM email e

LEFT JOIN user u1 ON e.send_id=u1.id

LEFT JOIN user u2 ON e.send_id=u2.id

WHERE u1.is_blacklist=0

AND u2.is_blacklist=0

GROUP BY e.date

ORDER BY e.date