I am reading http://www.django-rest-framework.org/api-guide/routers#usage and can't understand what a base_name is. Also i try to add a custom action and the router won't pick it up
I have this views.py
#authentication_classes((SessionAuthentication, TokenAuthentication))
#permission_classes((IsAuthenticated,))
class utente(CreateModelMixin, RetrieveAPIView, GenericViewSet, ViewSet):
model = MyUser
serializer_class = MyUserSerializer
def retrieve(self, request, *args, **kwargs):
self.object = MyUser.objects.get(
pk = request.user.pk
)
serializer = MyUserSerializerGET(self.object)
return Response(serializer.data)
#action(permission_classes=[IsAuthenticated])#POST action
def customaction(self, request):
return Response( None )
pass
and this urls.py
admin.autodiscover()
router_v1 = routers.DefaultRouter(trailing_slash=True)
router_v1.register(r'register', my_register, 'wtf' )
router_v1.register(r'utente', utente, 'wtf2' )
#router_v1.register(r'utente/customaction', utente.as_view({'post' : 'customaction'}) )
api_urls_v1 = router_v1.urls
api_urls = patterns('',
url(r'^v1/', include(api_urls_v1)),
)
urlpatterns = patterns('',
# Examples:
# url(r'^$', 'wecup.views.home', name='home'),
# url(r'^blog/', include('blog.urls')),
url(r'^admin/', include(admin.site.urls)),
url(r'^login/', 'rest_framework.authtoken.views.obtain_auth_token'),
url(r'^logout/', my_logout ),
url(r'^api/', include(api_urls)),
)
when i open http://127.0.0.1:8000/api/v1/
HTTP 200 OK Content-Type: application/json Vary: Accept Allow: GET, HEAD, OPTIONS
{
"register": "http://127.0.0.1:8000/api/v1/register/",
"utente": "http://127.0.0.1:8000/api/v1/utente/"
where is customaction?
}
You've got two different questions here, so I'll address them each separately.
Base Name
First, a base_name is simply the name the ViewSet will use when generating your named urls. By default, this will simply be your model or perhaps your queryset, though you may need to set it automatically if you've played with your ViewSet's get_queryset method.
If you don't implement your own url names, then the base_name will be used to implement them for you. For example, given that your Model is MyUser, your named urls would be something like 'myuser-list' or 'myuser-detail'.
Documentation, if interested, is here.
#action and custom methods
You're using a DefaultRouter, which allows you to access the API root view at http://127.0.0.1:8000/api/v1/, as you've shown. This root view only shows list views. Using #action creates a detail view. In your case, your customaction view can be found at ^utente/{pk}/customaction/$. It will not show up in the API root because it is not a list view.
General information on #action and custom methods can be found here.
Also, if for some reason you do want to make customaction a list-level view, you'll need to make some modifications. You could either string up a custom route yourself, without using the #action decorator (which is specifically for detail views). An example of that can be found here.
Your other option is to use the new-ish drf-extensions package. A discussion of using the package to implement collection level controllers in ViewSets can be found here.
Related
I have an Comment which have an Article foreign key (so Article have an "array" of comments). I need to build an url to fetch theese comments using article's pk, but when I am trying to do smth like "articles/int:article_pk/comments/" or "articles/{article_pk}/comments/" drf router crates static route with path "articles/{article_pk}/comments/". How can I implement getting a comments using article pk?
urls.py
router = DefaultRouter()
router.register('articles', articleAPI.ArticleAPI, basename='articles')
router.register('articles/comments', CommentAPI, basename='comments')
You can use this same url to get comments as well.
router.register('articles', articleAPI.ArticleAPI, basename='articles')
add a new method to Article ModelViewSet
#action(methods=['get'], detail=True,
url_path='comments', url_name='article-comments')
article = self.get_object()
serializer = CommentSerializer(queryset=article.comment_set.all(), many=True) # comment_set resembles the related name for article foreign key
return Response(serializer.data)
In postman hit the url articles/<article_id>/comments/ with GET method to get the comment list
Its not working because that is not how routers are intended to be used. You don't specify the key when registering, you use the viewset to define it. Read the docs at [1].
[1]: https://www.django-rest-framework.org/api-guide/routers/#simplerouter
A router can indeed not do that. You will need to capture the article_pk in the path and then work with two routers, so:
article_router = DefaultRouter()
article_router.register('articles', articleAPI.ArticleAPI, basename='articles')
comment_router = DefaultRouter()
comment_router.register('comments', CommentAPI, basename='comments')
urlpatterns = [
path('', include(article_router.urls)),
path('articles/<int:article_pk>/', include(comment_router.urls)),
]
I'm using REST in Django, And I couldn't understand what is the main difference between classic URL and instantiating DefaultRouter() for registering URL by ViewSet.
I have a model:
class Article(models.Model):
title = models.CharField()
body = models.TextField()
author = models.ForeignKey()
Serializing model like this:
from blog.models import Article
class ArticleSerializer(serializers.ModelSerializer):
class Meta:
model = Article
fields = ['title', 'body', 'author']
View Class:
from blog.models import Article
from rest_framework import viewsets
from .serializers import ArticleSerializer
class ArticleViewSet(viewsets.ModelViewSet):
serializer_class = ArticleSerializer
queryset = Article.objects.all()
and URLS:
router = DefaultRouter()
router.register(r'articles', ArticleViewSet)
urlpatterns = [
path('', include(router.urls)),
]
Is it possible to use classic URL in URLS.py instead of instantiating the object for a ViewSet like this:
urlpatterns = [
path('api/', 'views.someAPI'),
]
I just know HTTP method in ViewSet translate methods to retrieve, list and etc...
The Question is can we use traditional(Classic) URL style in this situation, Should we ?
Thanks for your help.
Well, in a nutshell as a django developer it is notorious how it is hard to deal with normal urls in django in some cases. Every now and again we get confused with the id type of the detail page that in some case are strings or integers with its regex, and so on.
For example:
urlpatterns = [
url(r'^(?P<content_type_name>[a-zA-z-_]+)$', views.content_type, name = 'content_type'),
]
# or
urlpatterns = [
url(r'^(?P<content_type_name>comics|articles|videos)$', views.content_type, name='content_type'),
]
Not mentioning that in almost every case its needed to have two urls like:
URL pattern: ^users/$ Name: 'user-list'
URL pattern: ^users/{pk}/$ Name: 'user-detail'
THE MAIN DIFFERENCE
However, using DRF routers the example above is done automatically:
# using routers -- myapp/urls.py
router.register(r"store", StoreViewSet, basename="store")
How django will understand it:
^store/$ [name='store-list']
^store\.(?P<format>[a-z0-9]+)/?$ [name='store-list']
^store/(?P<pk>[^/.]+)/$ [name='store-detail']
^store/(?P<pk>[^/.]+)\.(?P<format>[a-z0-9]+)/?$ [name='store-detail']
See how much job and headache you have saved with a line of code only?
To contrast, according to DRF documentation the routers is a type of standard to make it easy to declare urls. A pattern brought from ruby-on-rails.
Here is what the documentation details:
Resource routing allows you to quickly declare all of the common
routes for a given resourceful controller. Instead of declaring
separate routes for your index... a resourceful route declares them in
a single line of code.
— Ruby on Rails Documentation
Django rest framework documentation:
Some Web frameworks such as Rails provide functionality for
automatically determining how the URLs for an application should be
mapped to the logic that deals with handling incoming requests.
REST framework adds support for automatic URL routing to Django, and
provides you with a simple, quick and consistent way of wiring your
view logic to a set of URLs.
For more details follow the django rest framework documentation.
In the Django REST Framework Tutorial quickstart, views are added to the default router:
// project/urls.py
router = routers.DefaultRouter()
router.register(r'users', views.UserViewSet)
My project layout is
project/
stuffs/
I include the stuffs application in the default router:
router.register(r'stuffs', views.StuffViewSet)
The stuffs/ endpoint is then nicely listed in the API Root list of endpoints:
HTTP 200 OK
Allow: GET, HEAD, OPTIONS
Content-Type: application/json
Vary: Accept
{
"users": "http://localhost:8000/users/",
"groups": "http://localhost:8000/groups/",
"stuffs": "http://localhost:8000/stuffs/"
}
A Stuff object has a relationship to a owner model class and to filter for objects belonging to a certain owner, I intend to use a request such as:
stuffs/?owner_id=abc
In order to do so, and here is where I'm not sure if it is correct, I introduced an urls.py configuration in the stuffs app:
// stuffs/urls.py
urlpatterns = format_suffix_patterns([
path('stuffs/', stuffs_views.StuffList.as_view()),
path('stuffs/<int:pk>/', stuffs_views.StuffDetail.as_view()),
re_path(r'^stuffs/(?P<owner_id>.+)$', stuffs_views.StuffsList.as_view())
])
using the below StuffsList class based view:
class StuffList(generics.ListCreateAPIView):
serializer_class = StuffSerializer
def get_queryset(self):
queryset = Stuff.objects.all()
owner_id = self.request.query_params.get('owner_id')
if owner_id is not None:
queryset = queryset.filter(owner__owner_id=owner_id)
return queryset
I'm pretty sure this view is correct, because when I remove the View Set from the default router, the query works.
However, when the viewset is registered in the default router, when requesting stuffs/?owner_id=abc the filter is not applied.
I would like to have the viewset in the list of endpoints but at the same time be able to filter using the query parameter. How can I do that?
Why are you trying with short_id ? Anyway Try this:
def get_queryset(self):
queryset = Stuff.objects.all()
owner_id = self.request.query_params.get('owner_id')
if owner_id is not None:
queryset = queryset.filter(owner__short_id=int(owner_id))
return queryset
In Django Rest Framework you can simply generate documentation:
https://www.django-rest-framework.org/topics/documenting-your-api/#documenting-your-api
from rest_framework.documentation import include_docs_urls
urlpatterns = [
...
url(r'^docs/', include_docs_urls(title='My API title'))
]
Autogenerated documentation has request body nicely generated from serializer, nice documentation from docs but how to add authentication and permission classes information?
Some of my class-based views have custom authentication_classes and permission_classes and how to display information about them?
Look into include_docs_urls will see this:
def include_docs_urls(
title=None, description=None, schema_url=None, public=True,
patterns=None, generator_class=SchemaGenerator,
authentication_classes=api_settings.DEFAULT_AUTHENTICATION_CLASSES,
permission_classes=api_settings.DEFAULT_PERMISSION_CLASSES,
renderer_classes=None):
....
So, the solution is:
Setup the authentication_classes & permission_classes in settings.py
Sample authentication_classes:
# settings.py
REST_FRAMEWORK = {
'DEFAULT_AUTHENTICATION_CLASSES': (
'rest_framework_simplejwt.authentication.JWTAuthentication',
'rest_framework.authentication.SessionAuthentication', # <---- 1
),
'DEFAULT_PERMISSION_CLASSES': (
'common.permissions.IsDeveloper', # <---- 2 don't forget the ','
)
}
Note:
1. SessionAuthentication is needed because DRF Self describing APIs use session to keep user login
2. common.permissions.IsDeveloper is a custom permission class you can define your self and put the reference location as a string here, you can use default permission too.
Sample permission class:
# common/permissions.py
class IsDeveloper(permissions.BasePermission):
def has_permission(self, request, view):
user_groups = request.user.groups.values_list('name', flat=True)
return bool('Developer' in user_groups)
Then if you didn't login or login user not in 'Developer' Group, will come to:
I'm trying to display blog records for particular author using generic view:
urlpatterns = patterns('',
url(r'^blog/(?P<uid>[\d+])/$', ListView.as_view(
queryset=Blog.objects.filter(published=True, author=uid),
), name='blog_list'),
But I get NameError: name 'uid' is not defined
Is it possible to use urlconf named groups this way?
You need to create your own implementation of ListView like so:
class BlogListView(ListView):
model = Blog
def get_queryset(self):
return super(BlogListView, self).get_queryset().filter(
published=True, author__id=self.kwargs['uid'])
and then use it in your URLconf:
urlpatterns = patterns('',
url(r'^blog/(?P<uid>[\d+])/$', BlogListView.as_view(),
name='blog_list'),
The documentation for class-based generic views is, in my opinion, not quite up to scratch with the rest of the Django project yet - but there are some examples which show how to use ListView in this way:
https://docs.djangoproject.com/en/1.3/topics/class-based-views/#viewing-subsets-of-objects