django休息框架:router.DefaultRouter()url与自定义路径 [英] django rest framework: routers.DefaultRouter() url with custom path

查看:2782
本文介绍了django休息框架:router.DefaultRouter()url与自定义路径的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我想定义一个访问某个api的路径。到目前为止有什么工作(urls.py):

  router = router.DefaultRouter()
router.register(r 'test',views.TestViewSet)
urlpatterns = patterns('',
url(r'^ api /',include(router.urls)),

我想做的是添加一个新的ViewSet,它提供了test(urls.py)的子功能:

  router.register(r'test / add',views.TestNewViewSet)
pre>

但这不行。访问这个api时,我所得到的是一个404 Not Found错误。访问api时不会抛出异常。那么什么是错的?



任何帮助都不胜感激!

解决方案

尝试使用

  urlpatterns = patterns('',
url(r'^ api /',include(router。 url)),
url(r'^ test / add / $',TestNewViewSet.as_view(),name ='whatever'),


I want to define a path to access a certain api. What works so far (urls.py):

router = routers.DefaultRouter()
router.register(r'test', views.TestViewSet)
urlpatterns = patterns('',
    url(r'^api/', include(router.urls)),
)

What i would like to do is adding a new ViewSet which provides "subfunctionality" of test (urls.py):

router.register(r'test/add', views.TestNewViewSet)

But this does not work. When accessing this api, all i get is a "404 Not Found" error. No exceptions are thrown when accessing the api. So what is wrong?

Any help is appreciated!

解决方案

Try with

urlpatterns = patterns('',
url(r'^api/', include(router.urls)),
url(r'^test/add/$',  TestNewViewSet.as_view(), name='whatever'),

)

这篇关于django休息框架:router.DefaultRouter()url与自定义路径的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆