有沒有更好的方法來創建Django REST Web服務？

Question

我是Python Django的新手 - 對我的問題更進一步。

我創建一個REST Web服務，但我總是不滿意我的代碼，但我找不到更好的解決方案。是否真的有必要爲每個新的URL部分創建新的類？如果不是，我怎麼找到解決這個問題的更好方法？

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，你不必創建一個新的類新的URL，什麼您需要的是您的視圖中的新操作。

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 

檢查this question對@detail_route VS @list_route更多信息

然後，你可以通過一個請求調用它來movements/register