Question

我正在使用Django 1.8，我正在尝试为我的项目创建一个简单的feed。以下是创建Feed的feeds.py文件：

from django.contrib.syndication.views import Feed
from django.template.defaultfilters import truncatewords
from valueFact.models import ValueFactPost


class LatestPostsFeed(Feed):
    title = "Contributions from members"
    link = "/companies/"
    description = "New Contributions by members"

    def items(self):
        return ValueFactPost.published.all()[:5]

    def item_title(self, item):
        return item.title

    def item_description(self, item):
        return truncatewords(item.body, 30)

我的app url.py文件如下：

from django.conf.urls import url

from valueFact import views
from valueFact.feeds import LatestPostsFeed


urlpatterns = [
    url(r'^$',
        views.valueFactListView.as_view(),
        name='valueFact_list'),
    url(r'^(?P<year>\d{4})/(?P<post>[-\w]+)/$',
        views.valuefact_detail,
        name='valuefact_detail'),
    url(r'^(?P<fact_id>\d+)/share/$',
        views.valuefact_share,
        name='valuefact_share'),
    url(r'^feed/$', LatestPostsFeed(), name='post_feed'),
]

我将浏览器定向到网址'companies / feed'，并收到以下错误消息。请注意，我已经在主项目url conf文件中包含了所有这些应用程序URL，并且它正常工作（例如，当我输入127.0.0.1:8000/companies时，我得到了我想要的页面）

编辑：这是完整的跟踪

Environment:


Request Method: GET Request URL: http://127.0.0.1:8000/companies/feed/

Django Version: 1.8 Python Version: 3.5.0 Installed Applications: ('django.contrib.auth',  'django.contrib.contenttypes',  'django.contrib.sessions',  'django.contrib.messages',  'django.contrib.staticfiles',  'django.contrib.admin',  'django.contrib.sites',  'django.contrib.sitemaps',  'djangobower',  'rest_framework',  'stockData',  'accounts',  'functional_tests',  'valueFact') Installed Middleware: ('django.contrib.sessions.middleware.SessionMiddleware',  'django.middleware.common.CommonMiddleware',  'django.middleware.csrf.CsrfViewMiddleware',  'django.contrib.auth.middleware.AuthenticationMiddleware',  'django.contrib.auth.middleware.SessionAuthenticationMiddleware',  'django.contrib.messages.middleware.MessageMiddleware',  'django.middleware.clickjacking.XFrameOptionsMiddleware',  'django.middleware.security.SecurityMiddleware')


Traceback: File "/Users/djunh/Programming/viex/virtualenv/lib/python3.5/site-packages/django/core/handlers/base.py" in get_response
  132.                     response = wrapped_callback(request, *callback_args, **callback_kwargs) File "/Users/djunh/Programming/viex/virtualenv/lib/python3.5/site-packages/django/contrib/syndication/views.py" in __call__
  43.         feedgen = self.get_feed(obj, request) File "/Users/djunh/Programming/viex/virtualenv/lib/python3.5/site-packages/django/contrib/syndication/views.py" in get_feed
  174.                 request.is_secure(), File "/Users/djunh/Programming/viex/virtualenv/lib/python3.5/site-packages/django/contrib/syndication/views.py" in add_domain
  19.     if url.startswith('//'):

Exception Type: AttributeError at /companies/feed/ Exception Value: 'function' object has no attribute 'startswith'

提前致谢！

Answer 1

如果您未在模型上定义get_absolute_url(self)，则可能需要定义item_link

# item_link is only needed if NewsItem has no get_absolute_url method.
def item_link(self, item):
    return reverse('news-item', args=[item.pk])

来自docs：

要指定<link>的内容，您有两种选择。对于每个项目在 items（）中，Django首先尝试调用 item_link（）方法 Feed 类。与标题和描述类似，它是传递了一个参数 item 。如果该方法不存在， Django尝试在该对象上执行 get_absolute_url（）方法。 get_absolute_url（）和 item_link（）都应该返回商品的网址作为普通的Python字符串。与 get_absolute_url（）一样，结果如下 item_link（）将直接包含在网址中，您就是这样负责做所有必要的URL引用和转换方法本身内的ASCII。

Django Feed错误：'function'对象没有属性'startswith'

1 个答案: