Django URL，在URL中传递参数

Question

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'])

Answer 1

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

Answer 2

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