Python Django多对多查询的逻辑or返回重复的结果

我有这样多对多关系的模型：

class Contact(models.Model):
    name = models.TextField()
    address = models.TextField()

class Mail(models.Model):
    to = models.ManyToManyField(Contact, related_name='received_mails')
    cc = models.ManyToManyField(Contact, related_name='cced_mails')

我想获取给定电子邮件的“收件人”字段或“抄送”字段中的联系人集。让我们试试：

>>> Contact.objects.filter(received_mails__id=111)
[<Contact: fred@foo.com>]
>>> Contact.objects.filter(cced_mails__id=111)
[<Contact: joe@bar.com>]

---

当SQL返回所有匹配的记录时，Django尽职尽责地将它们映射到对象。您正在寻找的是

.distinct（）

queryset方法，该方法可以使SQL将所有重复的行折叠为一行。

成功了，谢谢。（很抱歉，我还不能投票支持你。）这将使用SELECT DISTINCT而不是SELECT更改查询。我不知道没有它你怎么能得到这么多的副本，但我眼前的问题已经解决了。你将为

mailshareapp\u联系人

×（

mailshareapp\u mail\u to

+

NULL

）×（

mailshareapp\u mail\u cc

+

NULL

）至少满足一个JOIN和一个WHERE子句的。顺便说一句，如果以上回答了你的问题，你可以随时将其标记为已接受的回答。

>>> Contact.objects.filter(Q(received_mails__id=111) | Q(cced_mails__id=111))
[<Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, <Contact: joe@bar.com>, '...(remaining elements truncated)...']

SELECT `mailshareapp_contact`.`id`, `mailshareapp_contact`.`name`,
`mailshareapp_contact`.`address` FROM `mailshareapp_contact`
LEFT OUTER JOIN `mailshareapp_mail_to`
ON (`mailshareapp_contact`.`id` = `mailshareapp_mail_to`.`contact_id`)
LEFT OUTER JOIN `mailshareapp_mail_cc`
ON (`mailshareapp_contact`.`id` = `mailshareapp_mail_cc`.`contact_id`)
WHERE (`mailshareapp_mail_to`.`mail_id` = 111
OR `mailshareapp_mail_cc`.`mail_id` = 111 )