Django：如何从管理器内部访问模型的实例？

发布于 2024-08-19 07:28:41 字数 482 浏览 4 评论 0原文

class SupercalifragilisticexpialidociousManager(models.Manager):
    # Sorry, I'm sick of Foo and Spam for now.
    def get_query_set(self, account=None):
        return super(SupercalifragilisticexpialidociousManager,
                     self).get_query_set().filter(uncle=model_thats_using_this_manager_instance.uncle)

我正在寻找的魔法是“uncle=model_thats_using_this_manager_instance.uncle”。看来我应该能够以某种方式做到这一点。我知道我可以说 self.model 来获取模型，但如何获取实例？

原文

class SupercalifragilisticexpialidociousManager(models.Manager):
    # Sorry, I'm sick of Foo and Spam for now.
    def get_query_set(self, account=None):
        return super(SupercalifragilisticexpialidociousManager,
                     self).get_query_set().filter(uncle=model_thats_using_this_manager_instance.uncle)

The magic I'm looking for is the "uncle=model_thats_using_this_manager_instance.uncle". It seems like I should be able to do this somehow. I know I could say self.model to get the model, but how to get the instance?

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凑诗 2024-08-26 07:28:41

当您使用管理器时，要求实例是没有意义的。管理器是类级属性 - 如果您尝试执行 foo.objects.all() ，其中 foo 是 Supercalifragilisticexpialidocious 的实例，您将明确收到错误：

AttributeError: Manager isn't accessible via Supercalifragilisticexpialidocious instances

It doesn't make sense to ask for an instance when you're using a manager. Managers are class-level attributes - if you try and do foo.objects.all() where foo is an instance of Supercalifragilisticexpialidocious, you will explicitly get an error:

AttributeError: Manager isn't accessible via Supercalifragilisticexpialidocious instances

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眼眸里的快感 2024-08-26 07:28:41

据我所知，您无法从管理器内部访问该模型。当管理者在整个桌子上操作时，这是没有意义的。

您应该在模型中

class Model(models.Model):
    # some attributes here
    def getAllRelativesWithSameUncle(self):
        return Model.objects.filter(uncle = self.uncle)

或在管理器中执行类似的操作：

class SupercalifragilisticexpialidociousManager(models.Manager):
    def getSelfRelativesFor(self, model):
        return self.get_queryset().filter(uncle=model)

As far as I know, you cannot access the model from inside a manager. It doesn't make sense as managers operate on the whole table.

You should do something like this in the model:

class Model(models.Model):
    # some attributes here
    def getAllRelativesWithSameUncle(self):
        return Model.objects.filter(uncle = self.uncle)

or in the manager:

class SupercalifragilisticexpialidociousManager(models.Manager):
    def getSelfRelativesFor(self, model):
        return self.get_queryset().filter(uncle=model)

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时光无声 2024-08-26 07:28:41

在像 object.lated_name.create() 这样的方法中，Django 会发送一个 hint 参数：

class UserQuerySet(QuerySet):
    def create(self, *args, **kwargs):
        print(self._hints)
        # >>> {'instance': <User: random-user>}
        print(self._hints.get('instance'))
        # >>> <User: random-user>

我现在使用的是 Django 1.11。

In methods like object.related_name.create() under the hood the Djangos sends a hint argument:

class UserQuerySet(QuerySet):
    def create(self, *args, **kwargs):
        print(self._hints)
        # >>> {'instance': <User: random-user>}
        print(self._hints.get('instance'))
        # >>> <User: random-user>

I'm using Django 1.11 nowadays.

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