特定网址的Django-rest-framework @detail_route [英] Django-rest-framework @detail_route for specific url

问题描述

我正在使用 Django-rest-framework == 3.3.2 和 Django == 1.8.8 。我有一个简单的 GenericView

 从rest_framework导入泛型
从rest_framework.decorators导入detail_route 
 
类MyApiView（generics.RetrieveAPIView）：
 serializer = MySerializer 
 def get（self，* args，** kwargs）：
 super（MyApiView，self）.get（* args，** kwargs）
 
 @detail_route（methods = ['post']）
 def custom_action（self，request）
 ＃做一些重要的事情
 return Response（）
  

如果我使用<$ django-rest-framework提供的c $ c> router ，但是我手动创建了所有网址，并希望对 detail_route 。

我想知道我是否可以做这样的事情：

  from django.conf.urls导入模式，myapi导入视图中的url 
 
 urlpatterns = pattern（
''，
 url（r'^ my-api / $' ，views.MyApiView.as_view（）），
 url（r'^ my-api / action $'，views.MyApiView.custom_action.as_view（）），
  

）

当然，第二个网址不起作用。

预先感谢。

解决方案

按照根据视图集文档中的示例，您可以将各个方法提取到视图中：

  custom_action_view = views.MyApiView.as_view（{ post： custom_action }）
  

然后，您可以自由地按常规方式进行路由：

  urlpatterns = [
 url（r'^ my-api / action $'，custom_action_view），
] 
  

我希望能帮上忙。

I'm using Django-rest-framework==3.3.2 and Django==1.8.8. I have a simple GenericView

from rest_framework import generics
from rest_framework.decorators import detail_route

class MyApiView(generics.RetrieveAPIView):
    serializer = MySerializer
    def get(self, *args, **kwargs):
        super(MyApiView, self).get(*args, **kwargs)

    @detail_route(methods=['post'])
    def custom_action(self, request)
        # do something important
        return Response()

This works fine if I use the router that django-rest-framework offers, however I'm creating all my urls manually and would like to do the same with the detail_route.

I wonder if it's possible for me to do something like this:

from django.conf.urls import patterns, url
from myapi import views
urlpatterns = patterns(
    '',
    url(r'^my-api/$', views.MyApiView.as_view()),
    url(r'^my-api/action$', views.MyApiView.custom_action.as_view()),

)

Of course this second url doesn't work. It's just an example of what I would like to do.

Thanks in advance.

解决方案

As per the example from the Viewsets docs, you can extract individual methods into views:

custom_action_view = views.MyApiView.as_view({"post": "custom_action"})

You're then free to route this as normal:

urlpatterns = [
  url(r'^my-api/action$', custom_action_view),
]

I hope that helps.

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