images = client.GetFeed('/data/feed/api/user/%s/album/%s?kind=photo' %
(GMAIL_USERNAME, str(album_name))).entry

Should give you the photos in the album with title album_name.  (ie.
drop the id from albumid in the feed url)

Harry

On Sep 11, 2:32 pm, "[email protected]"
<[email protected]> wrote:
> I don't know what happened to my previous post but I'm also very
> interested in this problem.
> I got it working with album ids but looking for something seo firendly
> like the solution descried by gurunars. (.net if possible)
> Any ideas?
>
> Many thanks
> K
>
> On Sep 9, 11:12 am, gurunars <[email protected]> wrote:
>
> > Hello,
>
> > I currently use Django framework and GdataAPI. Now to fetch a list of
> > photos for a particularalbumI have to use a view like this:
>
> > defalbum(request, album_identifier):
> >     client = gdata.photos.service.PhotosService()
> >     images = client.GetFeed('/data/feed/api/user/%s/albumid/%s?
> > kind=photo' % (GMAIL_USERNAME, str(album_identifier))).entry
> >     return render_to_response('picasa_album.html', {'images':images,})
>
> > In the code above album_identifier is actually an id of analbumkept
> > inside <gphoto:id> tag passed via URL from another page.
>
> > The question: is there any way to fetch the photos using a titlename
> > property of analbum? Or is it possible to get all the properties of a
> > particularalbumif its title is known? Currently I use iteration and
> > comparison in order to getalbum'sid if I know itsname. E.g.:
>
> > foralbumin albums:
> >     ifalbum.name.text == album_name:
> >         album_id =album.gphoto_id.text
> >         break
>
> > But this technique is crappy...
>
> > I want to access fetch the images viaalbum'stitle property because
> > this way I would be able to use more nice looking URLs on the site.
> > Not a freaky combination of digits but a readable string title
> > instead. Sth. like:
>
> > httP://example.com/photos/AlbumTitle
>
>
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