CODEWITHSUNDEEP
pythondjangorestswagger
David Brown
David Brown

Reputation: 938

Exclude URLs from Django REST Swagger

I have a few URLs that I want to exclude from my REST API documentation. I'm using Django REST Swagger and the only documentation I can find (https://github.com/marcgibbons/django-rest-swagger) doesn't really tell me much. There is the "exclude_namespaces" part of SWAGGER_SETTINGS in settings.py, but there is no real explanation or example of how to use this.

Simply put, I want to exclude any URLs from the docs that start with the following:

/api/jobs/status/
/api/jobs/parameters/

How could I go about doing this?

Thanks in advance for any help offered :P

Upvotes: 19

Views: 17342

Answers (7)

Evgene
Evgene

Reputation: 606

views.py

any view class

class ...ViewSet(viewsets.ModelViewSet):
    queryset = ....objects.all().order_by('-id')
    serializer_class = ...Serializer
    http_method_names = ['get', 'post', 'patch', 'delete'] # add or exclude

any function-based view

@api_view(['get']) # target field
def function(request):
    ...
    return Response(...)

Upvotes: 1

Vikram Ray
Vikram Ray

Reputation: 1146

A more flexible solution would be:

from django.contrib import admin
from django.urls import include, path
from rest_framework_swagger.views import get_swagger_view
urlpatterns = [
    path('admin/', admin.site.urls),
    path('users/', include('user.urls', namespace="user")),
    path('locations/', include('location.urls')),
    path('departments/', include('department.urls', namespace="department")),
    path('my_secret_api/', include('secret.urls', namespace="secret_api")),
]

to_exclude = ['secret_api',] # some more namespaces here
swagger_urls = [item for item in urlpatterns if hasattr(item,"namespace") and item.namespace not in to_exclude]
schema_view = get_swagger_view(title='Highky', patterns=swagger_urls)
urlpatterns += [
        path('api/docs/', schema_view),
]

urlpatterns will have all five paths, but swagger_urls will have four paths excluding secret_api.

All of your URLs and includes will continue to work as they were, except we are now passing our modified urlpatterns that we want to show in the Swagger docs. The checks will also cover the include where you don't specify a namespace (like in our case, where the namespace is not defined in the location).

Upvotes: 3

Ola Nguyen Van
Ola Nguyen Van

Reputation: 451

For all of those who found the above answer not helpful: I guess that "exclude_namespaces" doesn't work anymore in new versions of django swagger. I had almost the same problem (I didnt't want to show my internal apis in documentation) and the above solution didn't work for me. I've been searching for like an hour for a solution and finally found something helpful.

There are some attributes that you can pass to SchemaGenerator. One of them is urlconf. You can set it to be "yourproject.api.urls" and it will get only urls defined there! Of course, you have to make sure that all the urls that you want to exclude from your api documentation are not included there.

I hope that at least one person found my answer helpful ;).

A problem comes when you want to have many urls.py included in your api documentation. I don't know what should be done then. If anyone comes up with an answer to this new problem - feel free to comment my answer. thanks!

Upvotes: 19

Suyash Garg
Suyash Garg

Reputation: 101

With new version of django swagger, we don't need to create view to exclude some urls. Below code will disable test2 url.

from rest_framework_swagger.views import get_swagger_view
urlpatterns1 = [
    url(r'^', include(router.urls)),
    url(r'^test/', include('test.urls')),
    url(r'^test1/', Test2.as_view()),
]

schema_view = get_swagger_view(title='API Documentation', patterns=urlpatterns1)

urlpatterns = urlpatterns1 + [
    url(r'^docs/', schema_view),
    url(r'^test2/', Test2.as_view()),
]

Upvotes: 8

Jihoon Baek
Jihoon Baek

Reputation: 730

Ola's answer is correct. exclude_namespaces is no longer supported.

For finer control of the documentation, create your own schema view by using a function-based or class-based view. This can be useful if you want to produce documentation for specific URL patterns, or URL confs.

In your views.py, you can do the following:

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework.schemas import SchemaGenerator
from rest_framework_swagger import renderers

class SwaggerSchemaView(APIView):
    renderer_classes = [
        renderers.OpenAPIRenderer,
        renderers.SwaggerUIRenderer
    ]

    def get(self, request):
        generator = SchemaGenerator(title='Your API Documentation', urlconf='your_app.urls')
        schema = generator.get_schema(request=request)

    return Response(schema)

The above will only render documentation for the URLs that are specified in the urlconf argument of the SchemaGenerator. Also, don't forget to set up your urls.py as well:

from django.conf.urls import url
from views import SwaggerSchemaView

urlpatterns = [
    url(r'^api/v1/docs/$', SwaggerSchemaView.as_view(), name='docs'),
]

Upvotes: 6

Daria
Daria

Reputation: 753

For the newest version of drf-swagger you can defile url patterns in the schema generator.

For example: url_patterns = ( url(r'^api/v1/', include(router.urls, namespace='api')), ) generator = schemas.SchemaGenerator(title='Core API', patterns=url_patterns)

Upvotes: 4

Laurent Bristiel
Laurent Bristiel

Reputation: 6935

the namespaces to exclude are the one defined in your urls.py.

So for example, in your case:

urls.py:

internal_apis = patterns('',
                     url(r'^/api/jobs/status/',...),
                     url(r'^/api/jobs/parameters/',...),
                     )

urlpatterns = urlpatterns + patterns('',
              url(r'^', include(internal_apis, namespace="internal_apis")),
              ...
              )

and in your settings.py:

SWAGGER_SETTINGS = {
    "exclude_namespaces": ["internal_apis"],    #  List URL namespaces to ignore
}

This is well described in there

Upvotes: 19

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