Django Swagger starts failing when include is used in django urls

Question

I am using django rest_framework_swagger for my django project, everything was working fine but when I added some URLs with include method Swagger start giving me 500 internal server error.

I am not sure why this error is coming, I have checked but didn't find anything to fix this error.

I am using:

django 1.11.7
rest_framework_swagger 2.1.2
django rest framework 3.7.3

URLs

from django.conf.urls import url, include
from link_one.views import LinkOneViewSet
from link_two.views import LinkTwoViewSet

schema_view = get_swagger_view(title='My Project APIs')

urlpatterns = [

    url(r'^$', schema_view),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link1',
        LinkOneViewSet.as_view({'get': 'list'})),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link2',
        LinkTwoViewSet.as_view({'get': 'list'})),
    url(r'^foo/(?P<foo_id>\w+)/bar/(?P<bar_id>\w+)/link3',
        include('link_three.urls'))
]+ static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)

Error

    [25/Jan/2021 14:03:31] ERROR [django.request.exception:135] Internal Server Error: /
Traceback (most recent call last):
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\exception.py", line 41, in inner
    response = get_response(request)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\base.py", line 187, in _get_response
    response = self.process_exception_by_middleware(e, request)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\core\handlers\base.py", line 185, in _get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\views\decorators\csrf.py", line 58, in wrapped_view
    return view_func(*args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\django\views\generic\base.py", line 68, in view
    return self.dispatch(request, *args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 489, in dispatch
    response = self.handle_exception(exc)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 449, in handle_exception
    self.raise_uncaught_exception(exc)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\views.py", line 486, in dispatch
    response = handler(request, *args, **kwargs)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework_swagger\views.py", line 32, in get
    schema = generator.get_schema(request=request)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\generators.py", line 278, in get_schema
    links = self.get_links(None if public else request)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\generators.py", line 316, in get_links
    link = view.schema.get_link(path, method, base_url=self.url)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\inspectors.py", line 179, in get_link
    fields += self.get_serializer_fields(path, method)
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\schemas\inspectors.py", line 302, in get_serializer_fields
    serializer = view.get_serializer()
  File "C:\Users\myuser\conda_env\lib\site-packages\rest_framework\generics.py", line 112, in get_serializer
    return serializer_class(*args, **kwargs)
TypeError: 'list' object is not callable

that may be one option, but I can not upgrade django version. — Vikas Gautam, Jan 25 '21 at 08:47

score 0 · Accepted Answer · answered Jan 27 '21 at 07:40

After searching a lot and debugging, I have found a solution for this.

The solution is, don't use multiple serializer classes for a ViewSet.

In my one viewset I was doing this and this is what creating the problem.

class FooBarViewset(ModelViewSet): 
    serializer_class = [DefaultSerializer, BarSerializer, FooSerializer]

But I did not realize that this will cause the error.

Here is Fix that I am using

class FooBarViewset(ModelViewSet): 
    serializer_class = DefaultSerializer

You can also use the get_serializer method and map serializer class with an action, please check this answer Django rest framework, use different serializers in the same ModelViewSet

Django Swagger starts failing when include is used in django urls

1 Answers1