I spend some time for this one and propose next solution :

At first your model has only one  image. in this case  you dont need
to  add url() method . Now model is looks like


from django.db import models
from mptt.models import MPTTModel, TreeForeignKey
# Create your models here.
class BicycleAdItemKind(MPTTModel):
    def item_kind_image(self):
        return '<img align="middle" src="/media/%s" height="60px" />'
% self.image
    item_kind_image.allow_tags = True

    # Bicicleta completa, Componentes para bicicleta, Acessorios para ciclista
    n_item_kind      = models.CharField(max_length=50)
    parent           = TreeForeignKey('self', null=True,
                                      blank=True, related_name='children')
    description      = models.TextField(null=True, blank=True)
    image            = models.ImageField(upload_to='Images',
null=True, blank=True)
    date_inserted    = models.DateTimeField(auto_now_add=True)
    date_last_update = models.DateTimeField(auto_now=True)

    def __unicode__(self):
        return self.n_item_kind

    class MPTTMeta:
        order_insertion_by = ['n_item_kind']


Also view  should  be simple :

# Create your views here.
from django.views.generic import CreateView, ListView
from newmodel.models import  BicycleAdItemKind
from newmodel.form import CreatenewmodelForm

class newmodelListView(ListView):
    model = BicycleAdItemKind
    template_name = 'newmodel/index.html'

class newmodelAddView(CreateView):
    model = BicycleAdItemKind
    template_name = 'newmodel/create.html'
    form_class = CreatenewmodelForm
# please update form valid  method
    def form_valid(self, form, **kwargs):
       BicycleAdItemKind.objects.create()


 And  your  urls.py

from django.conf.urls import patterns, include, url
from newmodel.views import newmodelAddView, newmodelListView

# Uncomment the next two lines to enable the admin:
# from django.contrib import admin
# admin.autodiscover()

urlpatterns = patterns('',
    # Examples:
    # url(r'^$', 'andrelopes.views.home', name='home'),
    # url(r'^andrelopes/', include('andrelopes.foo.urls')),

    url(r'^newmodel/create$', newmodelAddView.as_view()),
    url(r'^newmodel/$', newmodelListView.as_view()),
    # Uncomment the admin/doc line below to enable admin documentation:
    # url(r'^admin/doc/', include('django.contrib.admindocs.urls')),

    # Uncomment the next line to enable the admin:
    # url(r'^admin/', include(admin.site.urls)),
)


 You  should create   two templates only it is not hard. I think that
form usage is over the top  and you can skip it

Many thanks,

Serge


+380 636150445
skype: skhohlov


2013/1/9 Sergiy Khohlov <[email protected]>:
>  I have no idea why you decided to investigate a bicycle...
>
> Please separate task  in two simple :
> 1) for editing object  use  url  /<objectname>/<id>/edit
> 2) for creating new object /<objectname>/new
>
>  Take a look  at generic view.
>
> Many thanks,
>
> Serge
>
>
> +380 636150445
> skype: skhohlov
>
>
> 2013/1/9 Tom Evans <[email protected]>:
>> On Wed, Jan 9, 2013 at 12:55 PM, Andre Lopes <[email protected]> wrote:
>>> Hi,
>>>
>>> I'm without a clue on how to get the sef.id when creating a new object
>>> in Django. I've created a question in StackOverflow. Can someone give
>>> me a clue on this subject?
>>>
>>> Please read the question:
>>> http://stackoverflow.com/questions/14234917/django-how-to-get-self-id-when-saving-a-new-object
>>>
>>> Any clues?
>>
>> Save the object first.
>>
>> Cheers
>>
>> Tom
>>
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