问Django Feed错误：“function”对象没有属性'startswith‘
EN

Stack Overflow用户

提问于 2016-04-04 05:10:05

回答 1查看 1.4K关注 0票数 1

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。下面是创建提要的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'),
]

我将浏览器定向到url 'companies/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！

python

django

feed

回答 1

Stack Overflow用户

发布于 2016-09-15 04:32:20

如果没有在模型上定义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，您有两个选项。对于 item ()中的每个项，Django首先尝试调用Feed类上的item_link()方法。它以类似于标题和描述的方式传递给它一个参数，item。如果该方法不存在，Django将尝试在该对象上执行get_absolute_url()方法。get_absolute_url()和item_link()都应该将项的URL作为普通的Python返回。与get_absolute_url()，一样，item_link()的结果将直接包含在URL中，因此您负责在方法本身内执行所有必要的URL引用和转换到ASCII。