On May 28, 6:30 am, sebey <[EMAIL PROTECTED]> wrote:
> from django.http import HttpResponse
> import feedparser
> from ubermicro.shows.models import show
>
> def show_page(request):
> """this is where we take what we need form the rss feeds in the
> data base"""
> query = show.objects.filter(show_feed__contains="http://";)
> podcast = feedparser.parse(query)
It seems that you are making feedparser parse an instance of a Django
ORM query. I think what you want to do is to make it parse a URL. May
be something like this:
for q in query:
podcast = feedparser.parse(q.show_feed)
if podcast.entries:
show_latest_title = podcast.entries[0].title
This will do it for all objects in your query. So you will have to
collect that list of titles etc. in a collection of some kind (list,
dict, etc.) and pass it on to your template.
> #show_about = podcast.feed.description
> show_latest_title = podcast.entries[0].title
The above statement assumes that there is at least one entry in the
feed. That may not be always true. So, you should consider testing
that first. Something like:
if podcast.entries:
show_latest_title = podcast.entries[0].title
> #show_latest = podcast.entries[0].description
>
> return HttpResponse(show_latest_title)
>
> this code is what I think its doing is that I am grabing the rss url
> then using feedparser (http://www.feedparser.org) to get rss element
> such as <description> and such but every time I try to do this I get
> this error
>
> Traceback (most recent call last):
> File "/Library/Frameworks/Python.framework/Versions/2.5/lib/python2.5/
> site-packages/django/core/handlers/base.py" in get_response
> 77. response = callback(request, *callback_args, **callback_kwargs)
> File "/Users/sebey/Sites/ubermicro/../ubermicro/shows/views.py" in
> show_page
> 10. show_latest_title = podcast.entries[0].title
>
> IndexError at /shows/
> list index out of range
> I am to django programing web dev etc. but I guess that the query I am
> useing is not correct so I what should I do thanks
-Rajesh
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