NoReverseMatch at /category/clothes/ Django class based views - django

Question

I have a django project that is close to an e-commerce wesite in terms of functionality.
There are four pages linked to one another. First page displays Categories, 2nd Subcategories, 3rd Product list and 4th Product detail and I'm using slugs to navigate.
ERROR
Reverse for 'product-list' with arguments '('', 'women-clothes')' and keyword arguments '{}' not found. 1 pattern(s) tried: ['category/(?P<category_slug>[-\\w]+)/(?P<subcategory_slug>[-\\w]+)/$']
Category to Subcategory linking code on the category_list.html is {{ category.name }} and on the views.py
class CategoryListView(ListView):
models = Category
template_name = 'products/category_list.html'
context_object_name = "Category list"
def get_queryset(self):
"""
Returns all categories.
"""
return Category.objects.get_queryset().all()
and urls.py
app_name = 'products'
urlpatterns = [
url(r'^$', CategoryListView.as_view(), name='categories'),
url(r'^(?P<category_slug>[-\w]+)/$', SubcategoryListView.as_view(), name='sub-category'),
url(r'^(?P<category_slug>[-\w]+)/(?P<subcategory_slug>[-\w]+)/$', ProductListView.as_view(), name='product-list'),
url(r'^(?P<category_slug>[-\w]+)/(?P<subcategory_slug>[-\w]+)/(?P<pk>\d+)/$', ProductDetailView.as_view(), name='product-detail'),]
The problem is linking subcategory_list.html to product_list. Since I need a category_slug and subcategory_slug to be pass to
{{ object.name }}.
I don't know how to implement this logic to using cbv. I want to pass category_slug since it is from a Category model and querying from Subcategory model.
views.py
class SubcategoryListView(ListView):
"""
Browse all products in the sub-catalogue.
"""
model = Subcategory
template_name = 'products/subcategory_list.html'
context_object_name = "Sub-Category list"
category_model = Category
def get_queryset(self):
"""
Returns all sub-categories.
"""
self.category = get_object_or_404(Category, category_slug = self.kwargs.get('category_slug'))
return Subcategory.objects.filter(category = self.category)
category.html which works.
{% for category in object_list %}
<div class="col-xs-12 col-md-12">
{{ category.name }}
<p>{{ category.category_slug }}</p>
</div>
{% endfor %}
subcategory.html
{% for object in object_list %}
<div class="col-xs-12 col-md-12">
{{ object.name }}
<p>subcategory_slug:{{ object.subcategory_slug }}</p>
</div>
{% endfor %}
How can get category_slug and pass it in the above view so as I can iterate on them on the template?

Answer 1

I don't really see what this has to do with CBVs. You haven't shown much of your template, but presumably you are iterating over subcategories and want to link to the individual list page for that subcategory. So, you just need to pass the slug and the category slug for the current subcategory in your loop.
This would be easier if you showed the rest of your template and your models, but assuming object is the subcategory, with a field called "subcategory_slug`, and the SubCategory model has an FK to Category:
{{ object.name }}

Answer 2

I was able to solve this error.
Changes
views.py
class SubcategoryListView(ListView):
"""
Browse all products in the sub-catalogue.
"""
model = Subcategory
template_name = 'products/subcategory_list.html'
context_object_name = "Sub-Category list"
category_model = Category
def get_queryset(self):
"""
Returns all sub-categories.
"""
self.category = get_object_or_404(Category, category_slug = self.kwargs.get('category_slug'))
return Subcategory.objects.filter(category = self.category)
def get_context_data(self, **kwargs):
"""
Returns self.category_slug needed
on the subcategory_list.html as a
one of the {% url %} slug params.
"""
context = super(SubcategoryListView, self).get_context_data(**kwargs)
context['categories'] = Category.objects.all()
context['category_slug'] = self.kwargs.get('category_slug')
return context
On the subcategory_list.html i changed object.category_slug to category_slug.
subcategory_list.html
{{ object.name }} .