Django: How do you access a model's instance from inside a manager?

Question

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?

score 7 · Answer 1 · answered Jan 25 '10 at 22:39

7

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

answered Jan 25 '10 at 22:39

Daniel Roseman

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Thanks Daniel. You're example helped me to understand the context in which managers exist. I still need to find a way to better handle this situation application-wide vs having to write special code hither and thither. – orokusaki Jan 26 '10 at 00:07

score 4 · Answer 2 · edited Jun 28 '16 at 14:24

4

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)

edited Jun 28 '16 at 14:24

mrmuggles

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answered Jan 25 '10 at 22:43

Felix Kling

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model method makes much more sense – Dmitry Shevchenko Jan 25 '10 at 22:57
I think you meant to say uncle=model_instance and (self, model_instance) rather than using the model itself as the filtering attribute. – orokusaki Jan 26 '10 at 00:05
Of course, thats why it is written in small letters ;) (I confess it could be misunderstood). Changed it.... – Felix Kling Jan 26 '10 at 00:10

Luan Fonseca · Accepted Answer · 2018-11-09T17:46:50.750

3

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.

edited Nov 09 '18 at 17:46

answered Nov 06 '18 at 22:38

Luan Fonseca

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score 0 · Answer 4 · answered May 13 '23 at 01:19

try self.model.

From the Docs

"Another thing to note is that Manager methods can access self.model to get the model class to which they’re attached." — https://docs.djangoproject.com/en/4.2/topics/db/managers/#adding-extra-manager-methods

What led to the answer: The manager knows what model to create when you call .create() from the manager.

Django: How do you access a model's instance from inside a manager?

4 Answers4