Use feedparser to sanitize HTML content for documents
Patch by: Sverre Rabbelier
from django.core import urlresolvers
from django.contrib.sitemaps import Sitemap
class GeoRSSSitemap(Sitemap):
"""
A minimal hook to produce sitemaps for GeoRSS feeds.
"""
def __init__(self, feed_dict, slug_dict=None):
"""
This sitemap object initializes on a feed dictionary (as would be passed
to `django.contrib.syndication.views.feed`) and a slug dictionary.
If the slug dictionary is not defined, then it's assumed the keys provide
the URL parameter to the feed. However, if you have a complex feed (e.g.,
you override `get_object`, then you'll need to provide a slug dictionary.
The slug dictionary should have the same keys as the feed dictionary, but
each value in the slug dictionary should be a sequence of slugs that may
be used for valid feeds. For example, let's say we have a feed that
returns objects for a specific ZIP code in our feed dictionary:
feed_dict = {'zipcode' : ZipFeed}
Then we would use a slug dictionary with a list of the zip code slugs
corresponding to feeds you want listed in the sitemap:
slug_dict = {'zipcode' : ['77002', '77054']}
"""
# Setting up.
self.feed_dict = feed_dict
self.locations = []
if slug_dict is None: slug_dict = {}
# Getting the feed locations.
for section in feed_dict.keys():
if slug_dict.get(section, False):
for slug in slug_dict[section]:
self.locations.append('%s/%s' % (section, slug))
else:
self.locations.append(section)
def get_urls(self, page=1):
"""
This method is overrridden so the appropriate `geo_format` attribute
is placed on each URL element.
"""
urls = Sitemap.get_urls(self, page=page)
for url in urls: url['geo_format'] = 'georss'
return urls
def items(self):
return self.locations
def location(self, obj):
return urlresolvers.reverse('django.contrib.syndication.views.feed', args=(obj,))