Django: Get model from string?

As of Django 1.11 to 4.0 (at least), it's AppConfig.get_model(model_name, require_ready=True)

As of Django 1.9 the method is django.apps.AppConfig.get_model(model_name). -- danihp

As of Django 1.7 the django.db.models.loading is deprecated (to be removed in 1.9) in favor of the the new application loading system. -- Scott Woodall

Found it. It's defined here:

from django.db.models.loading import get_model

Defined as:

def get_model(self, app_label, model_name, seed_cache=True):

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django.db.models.loading was deprecated in Django 1.7 (removed in 1.9) in favor of the the new application loading system.

Django 1.7 docs give us the following instead:

>>> from django.apps import apps
>>> User = apps.get_model(app_label='auth', model_name='User')
>>> print(User)
<class 'django.contrib.auth.models.User'>

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just for anyone getting stuck (like I did):

from django.apps import apps

model = apps.get_model('app_name', 'model_name')

app_name should be listed using quotes, as should model_name (i.e. don't try to import it)

get_model accepts lower case or upper case 'model_name'

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Most model "strings" appear as the form "appname.modelname" so you might want to use this variation on get_model

from django.db.models.loading import get_model

your_model = get_model ( *your_string.split('.',1) )

The part of the django code that usually turns such strings into a model is a little more complex This from django/db/models/fields/related.py:

    try:
        app_label, model_name = relation.split(".")
    except ValueError:
        # If we can't split, assume a model in current app
        app_label = cls._meta.app_label
        model_name = relation
    except AttributeError:
        # If it doesn't have a split it's actually a model class
        app_label = relation._meta.app_label
        model_name = relation._meta.object_name

# Try to look up the related model, and if it's already loaded resolve the
# string right away. If get_model returns None, it means that the related
# model isn't loaded yet, so we need to pend the relation until the class
# is prepared.
model = get_model(app_label, model_name,
                  seed_cache=False, only_installed=False)

To me, this appears to be an good case for splitting this out into a single function in the core code. However, if you know your strings are in "App.Model" format, the two liner above will work.

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The blessed way to do this in Django 1.7+ is:

import django
model_cls = django.apps.apps.get_model('app_name', 'model_name')

So, in the canonical example of all framework tutorials:

import django
entry_cls = django.apps.apps.get_model('blog', 'entry')  # Case insensitive

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2020 solution:

from django.apps import apps

apps.get_model('app_name', 'Model')

per your eg:

apps.get_model('people', 'Person')

per: Import Error :cannot import name get_model

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In case you don't know in which app your model exists, you can search it this way:

from django.contrib.contenttypes.models import ContentType 
ct = ContentType.objects.get(model='your_model_name') 
model = ct.model_class()

Remember that your_model_name must be lowercase.

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Another rendition with less code for the lazy. Tested in Django 2+

from django.apps import apps
model = apps.get_model("appname.ModelName") # e.g "accounts.User"

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I'm not sure where it's done in Django, but you could do this.

Mapping the class name to the string via reflection.

classes = [Person,Child,Parent]
def find_class(name):
 for clls in classes:
  if clls.__class__.__name__ == name:
   return clls

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Here is a less django-specific approach to get a class from string:

mymodels = ['ModelA', 'ModelB']
model_list = __import__('<appname>.models', fromlist=mymodels)
model_a = getattr(model_list, 'ModelA')

or you can use importlib as shown here:

import importlib
myapp_models = importlib.import_module('<appname>.models')
model_a = getattr(myapp_models, 'ModelA')

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