Inline.
2009/12/6 sleepjunk <sleepj...@gmail.com>:
> Thank you for the response. I don't believe I explained well enough
> exactly what I'm trying to do.
>
> I have table Video with a field vimeo_code. Each vimeo_code has its
> own XML file. The XML file I used in my views.py is just a random
> video I saw on vimeo. I need to replace the vimeo_code so the XML file
> I'm parsing looks like http://vimeo.com/api/v2/video/vimeo_code.xml
Ok, so just replace this line:
feed = urllib.urlopen("http://vimeo.com/api/v2/video/7875440.xml";)
with,
feed =
urllib.urlopen("http://vimeo.com/api/v2/video/%s.xml"; % v.vimeo_code)
This replaces the code in the URL with the current vimeo_code in the loop.
I think we're on the same page; give it a shot and see how it goes.
> Would it be a better idea to do this as a tag filter? Any additional
> information for me if I do it that way?
I'm sure someone will have a more concrete reason for doing it one way
or the other. Personally though, it's a matter of style. I prefer
having it in the view function because then I can see immediately that
extra http calls will be made when looking through my code. If I used
a tag filter, I would have to look at the view function to find the
template, then the template to see if custom tag filters were used,
then the .py files for the tag filters to see what it's doing.
Tag filters are pretty simple things; the Django docs link I gave
should be easy enough. Feel free to post questions if you get stuck
though.
> Thanks again, I appreciate your help.
> Matt
>
> On Dec 5, 10:07 pm, Sam Lai <samuel....@gmail.com> wrote:
>> 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,
>> seehttp://docs.djangoproject.com/en/dev/howto/custom-template-tags/
>>
>> HTH,
>>
>> Sam
>
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