django-rest-framework - 将 Django Rest Framework 的可浏览 API 与 APIViews 一起使用？

Question

如果我有这样的观点:

class MyAPIView(APIView):
    def get(self, request, name=None):
        return {"hello": name or "world"}

如何将其包含在生成的文档中？具体来说，如何将其包含在 API 根目录中，以便在访问“http://example.com/api/”时出现？

documentation includes an example带有描述的 APIView，但没有描述实际将其包含在 API 浏览器中的过程。

Answer 1

为了与路由器和 APIView 类或方法混合，使 API 根在 APIRoot View 中显示最少的代码 View ，我编写了一个自定义路由器，它扩展了 DefaultRouter 并覆盖 get_urls 和 get_api_root_view；它看起来如下:

from rest_framework import routers, views, reverse, response

class HybridRouter(routers.DefaultRouter):
    def __init__(self, *args, **kwargs):
        super(HybridRouter, self).__init__(*args, **kwargs)
        self._api_view_urls = {}

    def add_api_view(self, name, url):
        self._api_view_urls[name] = url

    def remove_api_view(self, name):
        del self._api_view_urls[name]

    @property
    def api_view_urls(self):
        ret = {}
        ret.update(self._api_view_urls)
        return ret

    def get_urls(self):
        urls = super(HybridRouter, self).get_urls()
        for api_view_key in self._api_view_urls.keys():
            urls.append(self._api_view_urls[api_view_key])
        return urls

    def get_api_root_view(self):
        # Copy the following block from Default Router
        api_root_dict = {}
        list_name = self.routes[0].name
        for prefix, viewset, basename in self.registry:
            api_root_dict[prefix] = list_name.format(basename=basename)
        api_view_urls = self._api_view_urls

        class APIRoot(views.APIView):
            _ignore_model_permissions = True

            def get(self, request, format=None):
                ret = {}
                for key, url_name in api_root_dict.items():
                    ret[key] = reverse.reverse(url_name, request=request, format=format)
                # In addition to what had been added, now add the APIView urls
                for api_view_key in api_view_urls.keys():
                    ret[api_view_key] = reverse.reverse(api_view_urls[api_view_key].name, request=request, format=format)
                return response.Response(ret)

        return APIRoot.as_view()

然后我用它作为 -

router = routers.HybridRouter()
router.register(r'abc', views.ABCViewSet)
router.add_api_view("api-view", url(r'^aview/$', views.AView.as_view(), name='aview-name'))
urlpatterns = patterns('',
    url(r'^api/', include(router.urls)),