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A Django app using 'feedparser' to fetch and parse a feed to render it from a template

License: MIT License

Python 96.73% HTML 3.27%

django-feedparser's Issues

Broken link to original repo

On pypi, this is listed as the homepage: https://github.com/emencia/django-feedparser

But that's a broken link. Could you fix this?

Can't unset the FEED_PLUGIN_TEMPLATE_CHOICES setting without hacking the code

For whatever reason, I can't override the default template without hacking the original code.
Ideally, it SHOULD be possible to just redefine the FEED_PLUGIN_TEMPLATE_CHOICES tuple in one's own settings file.

But that doesn't appear to work.

in the settings.py file:

> from django_feedparser.settings import *
> from cmsplugin_feedparser.settings import *
> FEED_PLUGIN_TEMPLATE_CHOICES = (
>     ('feedparser_plugin/default.html', 'Kiosk Default template'),
> )

Then in the shell:

>>> from django.conf import settings
>>> settings.FEED_PLUGIN_TEMPLATE_CHOICES
(('django_feedparser/basic_feed_renderer.html', 'Default template'),)
>>>

which is not the desired result.

Can't handle unicode string as URL (Python 3)

Traceback (most recent call last):
  File "/usr/local/lib/python3.4/dist-packages/django/core/handlers/base.py", line 132, in get_response
    response = wrapped_callback(request, *callback_args, **callback_kwargs)
  File "/usr/local/lib/python3.4/dist-packages/django/views/generic/base.py", line 71, in view
    return self.dispatch(request, *args, **kwargs)
  File "/usr/local/lib/python3.4/dist-packages/django/views/generic/base.py", line 89, in dispatch
    return handler(request, *args, **kwargs)
  File "/usr/local/lib/python3.4/dist-packages/django/views/generic/base.py", line 158, in get
    context = self.get_context_data(**kwargs)
  File "/usr/local/lib/python3.4/dist-packages/django_feedparser/views.py", line 47, in get_context_data
    context['feed_rendered'] = self.render_feed(feed_renderer, feed_url)
  File "/usr/local/lib/python3.4/dist-packages/django_feedparser/views.py", line 31, in render_feed
    return renderer().render(url, template=self.feed_template, expiration=self.feed_cache_expiration)
  File "/usr/local/lib/python3.4/dist-packages/django_feedparser/renderer.py", line 138, in render
    return render_to_string(template, self.get_context(url, expiration))
  File "/usr/local/lib/python3.4/dist-packages/django_feedparser/renderer.py", line 126, in get_context
    self._feed = self.get(url, expiration)
  File "/usr/local/lib/python3.4/dist-packages/django_feedparser/renderer.py", line 99, in get
    'id': hashlib.md5(url).hexdigest(),
TypeError: Unicode-objects must be encoded before hashing

Workaround is to pass the URL as a bytes string (b"")

sveetch / django-feedparser Goto Github PK

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django-feedparser's Issues

Broken link to original repo

Can't unset the FEED_PLUGIN_TEMPLATE_CHOICES setting without hacking the code

Can't handle unicode string as URL (Python 3)

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