I have been on vaction for a week, so I haven't had time to do more
coding before now. I think I understand what you have written in the
view, but I can't understand what to write in urls.py.
My views.py in my portfolio app folder looks like this:
from django.views.generic.list_detail import object_list
from myproject.portfolio.models import Project
from myproject.portfolio.models import Category
categories = Category.objects.all()
#Views
def category_view(request, cat_slug):
queryset = Project.objects.filter(category__slug = cat_slug)
context = {'category_slug': cat_slug}
return object_detail(request, queryset, extra_context =
context)
I have changed name with slug, but I tried with name and I got the same
error.
My urls.py looks like this:
urlpatterns = patterns('',
# For homepage:
(r'^$', 'django.views.generic.simple.direct_to_template',
{'template': 'homepage.html'}),
# Uncomment this for admin:
(r'^admin/', include('django.contrib.admin.urls')),
#Portfolio
(r'^portfolio/$', 'django.views.generic.list_detail.object_list',
dict(info_dict, template_name="portfolio/projects_list.html")),
(r'^portfolio/(?P<category_slug>[-\w]+)/$',
'myproject.portfolio.views.category_view'),
When I try "mydomain.com/portfolio/category_slug" (My category_slug is
"3d") I get the error:
TypeError at /portfolio/3d/
category_view() got an unexpected keyword argument 'category_slug'
So I think Im doing something wrong in the urls.py.
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