我是 Python Django 的新手 - 进一步回答我的问题。
我创建了一个 REST Web 服务，但我对我的代码非常不满意，但我找不到更好的解决方案。真的有必要为每个新的 URL 部分创建新的类吗？如果没有，我怎样才能找到更好的方法来解决这个问题？
urls.py
网址模式
= [
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")
...
请您参考如下方法:
不，您不必为新 url 创建新类，您需要的是 View 集中的新操作。
url(r'^movements/$', views.MovementDirection.as_view()),
View .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有关@detail_route 与@list_route 的更多信息
然后你可以通过请求movements/register
