I guess that video is a dictionary with one element. But dictionary.
Try video[0].id
On Feb 1, 1:06 pm, grimmus <graham.col...@gmail.com> wrote:
> When i write out just {{video}} it outputs the id of the object, but
> when i try {{video.id}} or {{video.title}} i get nothing.
>
> On Feb 1, 11:03 am, grimmus <graham.col...@gmail.com> wrote:
>
> > I am still experiencing issues with this.
>
> > Anyone any idea what i am doing wrong ?
>
> > Thanks !
>
> > On Jan 28, 3:28 pm, Daniel Roseman <dan...@roseman.org.uk> wrote:
>
> > > On Jan 28, 2:18 pm, grimmus <graham.col...@gmail.com> wrote:
>
> > > > Thanks for the reply. I am having some issues outputting the fields
> > > > now though.
>
> > > > It seems 4 objects are being returned alright but when i try and
> > > > output a value it's blank.
>
> > > > top_videos = {}
>
> > > > for i in xrange(1,6):
> > > > videos =Video.objects.filter(category=i).order_by('-
> > > > hit_count')
> > > > if videos:
> > > > top_videos[i] = videos[0]
>
> > > > t = loader.get_template('home/display.html')
> > > > c = RequestContext(request,{
> > > > 'top_videos': top_videos
> > > > })
> > > > return HttpResponse(t.render(c))
>
> > > > and in my template:
>
> > > > {% forvideoin top_videos %}
> > > > {{video.title}}
> > > > {% endfor %}
>
> > > > Any ideas what i am doing wrong ?
>
> > > For some reason, Gabriel's solution defined top_videos as a
> > > dictionary, so you would need to iterate through it as {% for
> > > category_id,videoin top_videos.items %}.
>
> > > A better way would be to define it as a list:
> > > top_videos = []
> > > for i in xrange(1,6):
> > > videos =Video.objects.filter(category=i).order_by('-
> > > hit_count')
> > > if videos:
> > > top_videos.append(videos[0])
>
> > > I must say, when I originally saw your post, I thought there must be
> > > an even simpler solution that gets to the topvideoin each category
> > > in a single query, using aggregation. But I've been unable to come up
> > > with the correct way of doing it.
> > > --
> > > DR.
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