What I was originally suggesting you do (which of course might not
turn out to be the best solution) is use feedparser to read feed urls,
and save those feeds into your database. That way, you wouldn't be
using feedparser at all in your views, just regular database queries.
If your models include Feed and FeedItem (FeedItem having a foreignkey
to a Feed), then feed parser will do something like this:
f = feedparser.parse('http://somefeed.url')
feed = Feed.objects.get(link=f.feed.link) # f.feed provides
information corresponding to one Feed instance
for item in f['entries']: # entries are all the items under a
particular feed
it, created = FeedItem.objects.get_or_create(title=item.title,
link=item.link, desc=item.description, date=item.date_parsed,
id=item.id, feed=feed)
Repeat the above for every url in your list of feed urls. Note: this
code goes in your standalone script running on a cron job, NOT in your
views or whatever other normal part of your django code base. That
will put RSS feed items into your database, your views.py pulls them
out again as per normal django methods:
feed = Feed.objects.get(link=somelinkvalue)
items = feed.feeditem_set.all()
return render_to_response(template, {'items':items})
That's how it ought it work, in my opinion. Note that it's late here,
and I've been drinking, so there may be some inaccuracies in the
above. I concur with the general consensus, that you need to start
with some simple CGI scripts and python basics, and move up from
there. That said, good luck!
E
On May 28, 2008, at 11:45 PM, sebey wrote:
>
> please much suggestions thank you
>
> On May 28, 4:37 pm, sebastian stephenson <[EMAIL PROTECTED]> wrote:
>> great finally some help thank you so much!
>> On 28 May 2008, at 15:20, Rajesh Dhawan wrote:
>>
>>
>>
>>
>>
>>> 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
>>
>> see ya
>>
>> sebey
> >
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