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Python rest_框架路由器模板中的链接_Python_Django_Django Rest Framework - Fatal编程技术网
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Python rest_框架路由器模板中的链接

python django django-rest-framework

Python rest_框架路由器模板中的链接,python,django,django-rest-framework,Python,Django,Django Rest Framework,我设置了rest\u框架，并将页面命名为api from . import views from rest_framework import routers from django.conf.urls import url from django.conf.urls import include router = routers.DefaultRouter() router.register(r'genres', GenreViewSet) router.register(r'blogs',B

我设置了rest\u框架，并将页面命名为
api

from . import views
from rest_framework import routers
from django.conf.urls import url
from django.conf.urls import include

router = routers.DefaultRouter()
router.register(r'genres', GenreViewSet)
router.register(r'blogs',BlogViewSet)

urlpatterns = [
    url(r'^api/',include(router.urls), name='api') #name is here???
]
但在模板中，我无法访问该页面

<a href="{% url 'api' %}">api</a>
您不能在“包含”路径上设置
名称
属性。

但是您可以设置
名称空间
属性来访问包含的视图：


{%url'命名空间：url\u名称“%}


django.urls.exceptions.NoReverseMatch: Reverse for 'api' not found. 'api' is not a valid view function or pattern name.