試圖導入TastyPie API，但我沒有找到頁面消息

我想創建一個Django網站，更重要的是，導入TastyPie API。但每次我在我的本地運行/用品/ API /條，我得到這個錯誤信息：試圖導入TastyPie API，但我沒有找到頁面消息

Page not found (404) 
Request Method:  GET 
Request URL: http://127.0.0.1:8000/articles/api/article 

Using the URLconf defined in django_test.urls, Django tried these URL patterns, in this order: 

    1. ^admin/ 
    2. ^accounts/login/$ 
    3. ^accounts/auth/$ 
    4. ^accounts/loggedin/$ 
    5. ^accounts/invalid/$ 
    6. ^accounts/logout/$ 
    7. ^accounts/register/$ 
    8. ^accounts/register_success/$ 
    9. ^articles/all/$ 
    10. ^articles/create/$ 
    11. ^articles/get/(?P<article_id>\d+)/$ 
    12. ^articles/like/(?P<article_id>\d+)/$ 
    13. ^articles/add_comment/(?P<article_id>\d+)/$ 
    14. ^articles/search/ 
    15. ^articles/api/article ^(?P<resource_name>article)/$ [name='api_dispatch_list'] 
    16. ^articles/api/article ^(?P<resource_name>article)/schema/$ [name='api_get_schema'] 
    17. ^articles/api/article ^(?P<resource_name>article)/set/(?P<pk_list>.*?)/$ [name='api_get_multiple'] 
    18. ^articles/api/article ^(?P<resource_name>article)/(?P<pk>.*?)/$ [name='api_dispatch_detail'] 

The current URL, articles/api/article, didn't match any of these.

這裏是我的urls.py文件，它位於我django_test/django_test目錄下：

from django.conf.urls import patterns, include, url 
from django.contrib import admin 
from django_test.api import ArticleResource 

article_resource = ArticleResource() 

urlpatterns = patterns('', 
         url(r'^admin/', include(admin.site.urls)), 
         url(r'^accounts/login/$', 'django_test.views.login'), 
         url(r'^accounts/auth/$', 'django_test.views.auth_view'), 
         url(r'^accounts/loggedin/$', 'django_test.views.loggedin'), 
         url(r'^accounts/invalid/$', 'django_test.views.invalid_login'), 
         url(r'^accounts/logout/$', 'django_test.views.logout'), 
         url(r'^accounts/register/$', 'django_test.views.register_user'), 
         url(r'^accounts/register_success/$', 'django_test.views.register_success'), 
         url(r'^articles/all/$', 'article.views.articles'), 
         url(r'^articles/create/$', 'article.views.create'), 
         url(r'^articles/get/(?P<article_id>\d+)/$', 'article.views.article'), 
         url(r'^articles/like/(?P<article_id>\d+)/$', 'article.views.like_article'), 
         url(r'^articles/add_comment/(?P<article_id>\d+)/$', 'article.views.add_comment'), 
         url(r'^articles/search/', 'article.views.search_titles'), 
         url(r'^articles/api/article', include(article_resource.urls)), 

)

，這是我api.py文件，它也位於我django_test/django_test目錄：

from tastypie.resources import ModelResource 
from tastypie.constants import ALL 
from article.models import Article 

class ArticleResource(ModelResource): 
    class Meta: 
     queryset = Article.objects.all() 
     resource_name = 'article'

我想，說不同的錯誤信息：

「對不起，尚未實現。請追加「格式= JSON？」你的URL「

我真的很困惑，我將不勝感激任何幫助，您可以給我謝謝

來源

2015-02-10 brown1001

看起來你需要傳遞一個RESOURCE_NAME參數，如：。 http://127.0.0.1:8000/articles/api/article/myresource – Selcuk 2015-02-10 16:51:54

我正在看TastyPie文檔，但我不知道'resource_name'是什麼意思。你的意思是把'resource_name'放在urls.py中？ – brown1001 2015-02-10 16:55:53

在urls.py你的最後一行應爲：。

url(r'^articles/api/', include(article_resource.urls)),

的article部分採用型號名稱tastypie自動生成你不應該包括在您的主urls.py

來源

2015-02-10 16:59:43 Selcuk

試圖導入TastyPie API，但我沒有找到頁面消息

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