Django - How to get admin url from model instance

Not trying to rip off @JosvicZammit, but using ContentType is the wrong approach here. It's just a wasted DB query. You can get the require info from the _meta attribute:

from django.urls import reverse

info = (model_instance._meta.app_label, model_instance._meta.model_name)
admin_url = reverse('admin:%s_%s_change' % info, args=(model_instance.pk,))

Just use this one liner that is also python 3 ready:

from django.urls import reverse

reverse('admin:{0}_{1}_change'.format(self._meta.app_label, self._meta.model_name), args=(self.pk,))

More on this in the django admin site doc, reversing admin urls.

This gives the same result as Josvic Zammit's snippet, but does not hit the database:

from django.urls import reverse
from django.db import models

class MyModel(models.Model):

    def get_admin_url(self):
        return reverse("admin:%s_%s_change" % (self._meta.app_label, self._meta.model_name), args=(self.id,))

This Django snippet should do:

from django.urls import reverse
from django.contrib.contenttypes.models import ContentType
from django.db import models

class MyModel(models.Model):

    def get_admin_url(self):
        content_type = ContentType.objects.get_for_model(self.__class__)
        return reverse("admin:%s_%s_change" % (content_type.app_label, content_type.model), args=(self.id,))

The self refers to the parent model class, i.e. self.id refers to the object's instance id. You can also set it as a property on the model by sticking the @property decorator on top of the method signature.

EDIT: The answer by Chris Pratt below saves a DB query over the ContentType table. My answer still "works", and is less dependent on the Django model instance._meta internals. FYI.