Django：通过外键字段的id进行查询的正确方法是什么？

Question

我有两个models：

class Organization(models.Model):
    title = models.CharField(max_length=100)

class Folder(models.Model):
    organization = models.ForeignKey("Organization",related_name='folders')
    title = models.CharField(max_length=50)

现在我想过滤folderby organization id。所以我尝试了：

• Folder.objects.filter(organization= 1)
• Folder.objects.filter(organization_id= 1)
• Folder.objects.filter(organization__id= 1)
• Folder.objects.filter(organization__pk= 1)
• Folder.objects.filter(organization= Organization.objects.get(id=1))

不管你相信与否，一切都会返回相同的结果。

那么有人知道通过外键字段的 id 进行查询的正确方法是什么吗？

更新

但是当尝试通过以下方式创建时folder：

Folder.objects.create(organization__id=1,title='hello')

出现错误：

Traceback (most recent call last):
  File "<console>", line 1, in <module>
  File "/home/suhail/.virtualenvs/heybadges/local/lib/python2.7/site-packages/django/db/models/manager.py", line 92, in manager_method
    return getattr(self.get_queryset(), name)(*args, **kwargs)
  File "/home/suhail/.virtualenvs/heybadges/local/lib/python2.7/site-packages/django/db/models/query.py", line 370, in create
    obj = self.model(**kwargs)
  File "/home/suhail/.virtualenvs/heybadges/local/lib/python2.7/site-packages/django/db/models/base.py", line 452, in __init__
    raise TypeError("'%s' is an invalid keyword argument for this function" % list(kwargs)[0])
TypeError: 'organization__id' is an invalid keyword argument for this function

但Folder.objects.create(organization_id=1,title='hello')效果很好。

Answer 1

Django 文档说你应该Folder.objects.filter(organization__pk=1)在大多数情况下使用。