我想在Django中提出一个URL定义来支持这个
/类别/ categorie_slug /产品/ product_slug
其中，categorie_slug为Unique，而product_slug对于父类别是唯一的。
我的观点：
"""
Listviews
"""
class CategorieList(ListView):
context_object_name = "categorie_list"
model = Categorie
queryset = Categorie.objects.order_by('-created_at')
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super(CategorieList, self).get_context_data(**kwargs)
# Add in a QuerySet of all Products
context['product_list'] = Product.objects.all()
return context
class ProductList(ListView):
context_object_name = "product_list"
model = Product
def get_queryset(self):
categorie = self.kwargs['categorie']
return Product.objects.filter(categorie=categorie)
"""
Detailviews
"""
class CategorieDetail(DetailView):
context_object_name = "categorie"
model = Categorie
def get_context_data(self, **kwargs):
# Call the base implementation first to get a context
context = super(CategorieDetail, self).get_context_data(**kwargs)
categorie = self.object
# Add in a QuerySet of all posts for the given estimates
context['product_list'] = Product.objects.filter(categorie=categorie)
return context
class ProductDetail(DetailView):
context_object_name = "product"
model = Product
我的网址格式：
urlpatterns = patterns('',
url(r'^$', login_required(CategorieList.as_view()),name="categories"),
url(r'^.*categories/$', login_required(CategorieList.as_view()),name="categories"),
url(r'^.*categories/(?P<slug>[\w-]+)/$', login_required(CategorieDetail.as_view()), name='categorie'),
url(r'^.*categories/(?P<slug>[\w-]+)/products/$', login_required(Productlist.as_view()), name="products"),
url(r'^.*categories/(?P<slug>[\w-]+)/products/(?P<slug>[\w-]+)/$', login_required(ProductDetail.as_view()), name="product"),
)
问题在于我必须将categorie-slug作为第二个参数添加到我的Productdetail中，因为两个slug都使产品独一无二。