How to get an app name using python in django

Question

If you are in the view and want to retrieve the app name using Python ( the app name will be used for further logic ), how would you do it ?

Answer 1

You could do:

from django.core.urlresolvers import resolve

....

resolve(request.path).app_name

See How to get current application in Django and resolve()

EDIT: You can now use request.resolver_match.app_name which avoids resolving a second time and avoids an import. Do it this way:

request.resolver_match.app_name

Answer 2

Another way to do it is get the current object or use self. see bellow

obj.__module__.split('.')

This will return a list with the object name split up by the '.'

Answer 3

You can get application name from model: Book._meta.app_label.

I've found in django/contrib/admin/widgets.py:

class RelatedFieldWidgetWrapper(forms.Widget):
    ...
    def get_context(self, name, value, attrs):
        from django.contrib.admin.views.main import IS_POPUP_VAR, TO_FIELD_VAR
        rel_opts = self.rel.model._meta
        info = (rel_opts.app_label, rel_opts.model_name)
        ...
    ...

Answer 4

__package__

or

__package__.rsplit('.', 1)[-1]

should be the easiest way. Second converts a.b.c to c.

Answer 5

maybe this can help you..

from django.core import urlresolvers
from django.contrib.contenttypes.models import ContentType

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

url will return you all adress and you can parse it for your app and model...

Answer 6

so to sum up, If you're looking for the app_name of where your code is written, then use this:

app_name = __package__

However, if you need the app_name of where the object is being called (which might be in an app other than coding app), then use this:

import sys
from django.urls import resolve

app_name = sys.modules[resolve(request.path_info).func.__module__].__package__

Answer 7

This solution not very simple, but it works

import sys

sys.modules[resolve(request.path_info).func.__module__].__package__