Django URL，在URL中传递参数

Question

我想创建如下网址：

/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

正如@Umair所说，您要将这些键作为URL查询参数进行传递，因此不必在URLPATTERNS提及它

对于您的情况，要获取通过URL传递的数据，请遵循以下代码段

#urls.py
urlpatterns = [
                  path('admin/', admin.site.urls),
                  url(r'^api/foodfeed/', 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): print(request.GET) # print all url params print(request.GET['keywords']) print(request.GET['mood'])
        # etc

Answer 2

这些keywords ， mood ， location等是query params您不应在url中包括这些query params ，而应通过request.query_params访问它们

参考： http : //www.django-rest-framework.org/api-guide/requests/#query_params