题解 | 异常的邮件概率

SELECT date,round(avg(type = 'no_completed'),3) p
FROM(   
    SELECT a.type,a.date
    FROM email AS a 
    INNER JOIN user AS b ON a.send_id = b.id AND b.is_blacklist = 0
    INNER JOIN user AS c ON a.receive_id = c.id AND c.is_blacklist = 0
) AS d
GROUP BY date
ORDER BY date ASC

好多题解啊，随便看看吧