Django Feed 错误："函数"对象没有属性"startswith"

Question

我正在使用Django 1.8，并试图为我的项目创建一个简单的提要。这是创建提要的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)

我的应用程序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'),
]

我将浏览器指向url"company/feed"，并得到以下错误。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的url，并且它正常工作（例如，当我键入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'

提前感谢！

Answer 1

如果您没有在模型上定义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])

来自文档：

要指定<link>的内容，有两个选项。对于每个项目在items（）中，Django首先尝试在馈送类。与标题和描述类似向它传递了一个参数，项。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（）都应返回项目的url作为一个普通的Python字符串。与get_absolute_url（）一样item_link（）将直接包含在URL中，因此您负责进行所有必要的URL引用和转换方法本身内部的ASCII。