See inline.

2009/12/6 sleepjunk <sleepj...@gmail.com>:
> Hello :) I am working on a small video site that allows a user to add
> a video to the site by submitting a Vimeo.com or Youtube.com URL. I'm
> trying to display thumbnails without hosting them. For Vimeo videos I
> need to parse an XML file to find the thumbnail location.
>
> On my homepage I would like to display the 5 newest videos. As of now
> I'm displaying every video in the database on the homepage. Each video
> will have a different thumbnail and XML file to parse. As of right now
> my "main_page" in my views.py file looks like this:
>
> def home_page(request):
>        template = get_template('home_page.html')
>        videos = Video.objects.all()

          videos_context = [ ]
          for v in videos:
>             feed = urllib.urlopen("http://vimeo.com/api/v2/video/7875440.xml";)
>             tree = ET.parse(feed)
>             root = tree.getroot()
>             vimeo_thumbnail = root.findtext("video/thumbnail_medium")
               videos_context.append( { 'video' : v, 'thumbnail_url' :
vimeo_thumbnail } )

>        variables = Context({
>                'videos' : videos_context,
>        })
>        output = template.render(variables)
>        return HttpResponse(output)
>
> How can I parse a different XML file for every video? Thank you much.
> Again, I am new and could be going about this completely wrong and
> gladly accept any feedback you have.

You could try something like the modified view above. It loops through
every video, gets the thumbnail URL, creates an object containing the
video model instance and the thumbnail URL, and adds it to the list
videos_context.

You could then access them in your template like this (assuming
there's a name field in your Video model):

{% for v in videos %}
<img src="{{v.thumbnail_url}}">{{ v.video.name }}
{% endfor %}

The alternate way of doing this is to make a tag filter, see
http://docs.djangoproject.com/en/dev/howto/custom-template-tags/

HTH,

Sam

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