I want to create url like:
/api/foodfeeds/?keywords=BURGER,teste&mood=happy&location=2323,7767.323&price=2
urls.py
urlpatterns = [
path('admin/', admin.site.urls),
url(r'^api/foodfeed/(?P<keywords>[0-9.a-z, ]+)/(?P<mood>[0-9.a-z, ]+)/(?P<location>[0-9]+)/(?P<price>[0-9]+)/$', backend_views.FoodfeedList.as_view()),
]+ static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)
views.py
class FoodfeedList(APIView):
# permission_classes = (permissions.IsAuthenticated,)
def get(self,request,keywords,mood,location,price):
print(request.GET['keywords'])
As @Umair said, you're passing those keys as URL query parameters , so you don't have to mention it in URLPATTERNS
In your case, to get the data you're passing through the URL, follow the below code snippet
#urls.py
urlpatterns = [
path('admin/', admin.site.urls),
url( backend_views.FoodfeedList.as_view()),
] + static(settings.STATIC_URL, document_root=settings.STATIC_ROOT)
#views.py
class FoodfeedList(APIView):
# permission_classes = (permissions.IsAuthenticated,)
# etc
Those keywords
, mood
, location
, etc are query params
you should not include those in url, rather you should access them via request.query_params
Reference : http://www.django-rest-framework.org/api-guide/requests/#query_params
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