我有一个django在功能方面接近电子商务网站的项目。NoReverseMatch at/category/clothes/Django class based views
有四页链接彼此。第一页显示分类,第二小类,第3 产品列表和第4 产品详细和我使用蛞蝓导航。
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]+)/$']
分类到子类别在category_list.html链接码是<a href="{% url 'products-app:sub-category' category.category_slug %}">{{ category.name }}</a>
并且在views.py
class CategoryListView(ListView):
models = Category
template_name = 'products/category_list.html'
context_object_name = "Category list"
def get_queryset(self):
"""
Returns all categories.
"""
return Category.objects.get_queryset().all()
和urls.py
app_name = 'products'
urlpatterns = [
url(r'^$', 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'),]
问题在于链接subcategory_list.html到product_list。因为我需要一个category_slug和subcategory_slug要传递给
<a href="{% url 'products-app:product-list' category_slug subcategory_slug %}">{{ object.name }}</a>
。
我不知道如何实现这个逻辑来使用cbv。我想通过category_slug因为它是从类别模型和查询从子类别模型。 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其中工程。
{% for category in object_list %}
<div class="col-xs-12 col-md-12">
<a href="{% url 'products-app:sub-category' category.category_slug %}">{{ category.name }}</a>
<p>{{ category.category_slug }}</p>
</div>
{% endfor %}
subcategory.html
{% for object in object_list %}
<div class="col-xs-12 col-md-12">
<a href="{% url 'products-app:product-list' object.category_slug object.subcategory_slug %}">{{ object.name }}</a>
<p>subcategory_slug:{{ object.subcategory_slug }}</p>
</div>
{% endfor %}
怎样才能得到category_slug,并通过它上面视图所以,我可以对他们进行迭代的模板?
我已经更新了这两个文件。 @Daniel Roseman –