On Saturday, 14 February 2015 04:28:40 UTC, inoyon artlover KLANGRAUSCH
wrote:
>
>
> urlpatterns = patterns('',
> url(r'^admin/', include(admin.site.urls)),
> url(r'^(?P)$',List.as_view()),
> )
>
That regex will never match anything. You need to give it some character
classes to match:
models.py
slug = models.SlugField(verbose_name='Slug', unique=True)
@models.permalink
def get_absolute_url(self):
return ('myapp:List', (), {'slug': self.slug})
admin.py
from .models import List01
class ArticleAdmin(admin.ModelAdmin):
prepopulated_fields = {"slug": ("name",)}
admin.s
#views
def tests(request):
acept = get_object_or_404(Modeltest, released=True)
context = {'keys': acept} # [1]
return render(request, 'temp_test.html', context)
#template
{% for key in keys %} # [1]
{{key.title}}
{{key.name}}
{{key.number}}
{{key.etc}}
{% endfor %}
I am developing a site on django. I want to hit a api url. Please suggest
me. Its realy very urgent. Thanks
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seem to be having issues with it one question if anyone has if you make a
new project from scratch do you still need to use the virtualenv to create
a project
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On 14/02/2015 8:06 PM, Kapil Tomer wrote:
I am developing a site on django. I want to hit a api url. Please
suggest me. Its realy very urgent. Thanks
http://www.django-rest-framework.org/tutorial/quickstart/
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Do you want code in your Django project to contact an API provided by
another site in order to display information on your site? Or are you
writing an API to provide services for end users?
If the latter, I would recommend the DRF in the link provided in the
previous response by Mike.
-James
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