题解 | #异常的邮件概率#

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