django redirect url with parameter

Question

when user click make thumbnailbutton ,it will call def thumbin the view and then render to http://127.0.0.1:8000/upload/thumb

I want to change to redirect to the url with the id number in database(Item.objects.all()) like : http://127.0.0.1:8000/upload/thumb/123

But not get it. Please help me.Thank you very much.

My code:

urls.py

urlpatterns = patterns('', 
    url(r'^$', views.background, name='background'),
    url(r'^thumb/(?P<result>\d+)$', views.thumb, name='thumb'),

views.py

def thumb(request,result):
    if request.method=="POST":
        photoid = request.POST['photoid']
        photowidth = request.POST['photowidth']
        item=Item.objects.filter(id=photoid) 
        return redirect(reverse('imageupload:thumb',kwargs={'result':photoid,'item':item }))

return HttpResponseRedirect(reverse('imageupload:background'))

templates:

  <form action="{% url 'imageupload:thumb'  i.id %}" method="POST" id="create_post">

Perhaps you mean to use kwargs={'item' : item} instead of args. Note, however, that if you're using named parameters in your url definition then your kwarg name will need to match. aka: kwargs={'result' : item} — RutledgePaulV
– RutledgePaulV, Commented Dec 9, 2014 at 3:08
I think in general, however, you might be a bit confused since the code doesn't seem to make a lot of sense and there are several problems. Rather than posting a bunch of code can you specify what it is you're trying to do? — RutledgePaulV
– RutledgePaulV, Commented Dec 9, 2014 at 3:10

shellbye · Accepted Answer · 2014-12-09 03:31:02Z

4

When you use named parameter, you need to do like this:

return redirect(reverse('imageupload:thumb',kwargs={'result':item}))

And in your forms.py, you also need to modify your action to this:

action="{% url 'imageupload:thumb' result %}"

You can also access result in views.py like this:

def thumb(request, result):
    print result

answered Dec 9, 2014 at 3:31

shellbye

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