Class Feeds DRY TemplateTag
I'm using the Django Feeds Framework and it's really nice, very intuitive and easy to use. But, I think there is a problem when creating links to feeds in HTML. For example: <link rel="alternate" type="application/rss+xml" title="{{ feed_title }}" href="{{ url_of_feed }}" /> Link's `HREF` attribute can be easily found out, just use `reverse()` But, what about the `TITLE` attribute? Where the template engine should look for this? Even more, what if the feed is build up dinamically and the title depends on parameters (like [this](http://docs.djangoproject.com/en/dev/ref/contrib/syndication/#a-complex-example))? This is the solution I came up with. However, as you can see, there is some caveats: * Requires Django 1.2 Class Feeds, don't know exactly how to do this with the old way of feeds. * If the feed class uses the request object, the `request` [context processor](http://docs.djangoproject.com/en/dev/ref/templates/api/#django-core-context-processors-request) must be configured, since `None` is passed if it isn't present in the context. * There's an oddity with Feed.__get_dynamic_attr(). The Feed subclass instance doesn't have this method; instead, it appears with another name. Don't know how to figure the name out at runtime... I've posted this problem on [StackOverflow](http://stackoverflow.com/questions/2784659/django-dry-feeds), but didn't get a better answer.
- Feeds DRY templatetag HTML LINK