有没有更好的方法来创建Django REST Web服务？

Question

我是Python Django的新手 - 对我的问题更进一步。

我创建一个REST Web服务，但我总是不满意我的代码，但我找不到更好的解决方案。是否真的有必要为每个新的URL部分创建新的类？如果不是，我怎么找到解决这个问题的更好方法？

urls.py

urlpatterns的

= [ 
    url(r'^admin/', admin.site.urls), 
    url(r'^movement/$', views.MovementDirection.as_view()), 
    url(r'^movement/on/$', views.On.as_view()), 
    url(r'^movement/off/$', views.Off.as_view()), 
    url(r'^movement/stop/$', views.Stop.as_view()), 
    url(r'^movement/forward/$', views.Forward.as_view()), 
    url(r'^movement/backwards/$', views.Backward.as_view()), 
] 

urlpatterns = format_suffix_patterns(urlpatterns) 

views.py

... 
#movement/forward 
class Forward(APIView): 
    def get(self, request): 
     print("FORWARD") 
     # engine control 
     return Response("Forward") 

#movement/backwards 
class Backward(APIView): 
    def get(self, request): 
     print("BACKWARD") 
     # engine control 
     return Response("Backward") 
... 

Answer 1

不，你不必创建一个新的类新的URL，什么您需要的是您的视图中的新操作。

url(r'^movements/$', views.MovementDirection.as_view()), 

views.py

class MovementDirection(ModelViewSet): # please notice the class 
    @list_route 
    def register(self, request): 
     # you can make it on a whole endpoint 
     # or on a single instance by using @detail_route 

检查this question对@detail_route VS @list_route更多信息

然后，你可以通过一个请求调用它来movements/register