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
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.
while practicing on Django I have faced the following problem of Identical URL Pattern. I have tried but fail to solve.
My project has one apps and the models are
class Category(models.Model):
title = models.CharField(max_length=150)
slug = models.SlugField(unique=True,blank=True, max_length=255)
class Post(models.Model):
title = models.CharField(max_length=200)
cats = models.ForeignKey(Category, on_delete=models.CASCADE)
slug = models.SlugField(unique=True,blank=True, max_length=255)
My urls.py
from django.contrib import admin
from django.urls import path
from post import views as apps
urlpatterns = [
path('admin/', admin.site.urls),
path('<slug:slug>/', apps.categoryView, name='category'),
path('<slug:slug>/',apps.PostView, name='calc_detail'),
]
Problem:
When I Put '/' in the second line of the urlpattern for Category View, category View works but post View doesn't work (404).
If remove '/' urlpattern for CategoryView , then post view Works but Category Views shows 404.
How should I configure these urls. I am using function based views
There's really no way to handle these except using different patterns or handling both in the same view, I would suggest different patterns:
from django.contrib import admin
from django.urls import path
from post import views as apps
urlpatterns = [
path('admin/', admin.site.urls),
path('pattern1/<slug:slug>/', apps.categoryView, name='category'),
path('pattern2/<slug:slug>/',apps.PostView, name='calc_detail'),
]
Also have you thought what would happen if a post and a category ended up having the same slug? Best to have different patterns.
About putting '/' making some url work and not it is because django appends slashes by default to each url, so if you really want to do this (NOT RECOMMENDED) you may set APPEND_SLASH = False in your setting and have one url with a slash and other without.
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.
I am new to Django and would greatly appreciate your help.
I have the below mentioned piece of code from an older book about Django. However, django.views.generic.list_detail has been deprecated. Can someone tell me how I could re-write this code with django.views.generic.list.ListView?
from django.conf.urls import patterns, include, url
from cmsproject.cms.models import Story
info_dict = {'queryset': Story.objects.all(), 'template_object_name': 'story'}
urlpatterns = patterns('django.views.generic.list_detail',
url(r'^(?P<slug>[-\w]+)/$', 'object_detail', info_dict, name="cms-story"),
url(r'^$', 'object_list', info_dict, name="cms-home"),
)
Assuming all you want to do is fetch a list of Story model objects, this is one way to write your views.py and urls.py:
In views.py:
from django.views.generic.list import ListView, DetailView
from cmsproject.cms.models import Story
class StoryListView(ListView):
model = Story
template_name = "cms/story_list.html"
class StoryDetailView(DetailView):
model = Story
template_name = "cms/story_detail.html"
template_name depends on where in the project you placed your html files. By setting model = Story, ListView will fetch Story.objects.all(). To customize, filter, add context etc., you can override any of the methods that your class based view inherits from its parent view (ex. in StoryListView you can override ListView methods).
In urls.py
from django.conf.urls import patterns, url
from cmsproject.cms.views import StoryDetailView, StoryListView
urlpatterns = patterns('',
url(r'^(?P<slug>[-\w]+)/$', StoryDetailView.as_view(), name="cms-story"),
url(r'^$', StoryListView.as_view(), name="cms-home"),
)
Think of urls.py as a mapping between the url and the View object. Defining name allows you to refer/link to other views by including name as a parameter to url template tag in the templates.
Some Extremely Useful References:
Effective Django - Class Based Views
CCBV List View
Django Project Generic Display Views
Built-in template Tags and filters -> look under url