How can use slug for all urls in django without anything before or after?
I want all djaango urls use slug field without any parameter before or after, by default just one url can use this metod
ex:
path('<slug:slug>', Article.as_View(), name="articledetail"),
path('<slug:slug>', Category.as_View(), name="category"),
path('<slug:slug>', Product.as_View(), name="productdetail"),
mysite .com/articleslug
mysite .com/categoryslug
mysite .com/productslug
How can I do it? Thank you
It will always trigger the Article
view, regardless if there is an Article
for that slug. You thus should make the URL patterns non-overlapping such that the other views can be triggered, for example with:
path('article/<slug:slug>/', Article.as_View(), name="articledetail"),
path('category/<slug:slug>/', Category.as_View(), name="category"),
path('product/<slug:slug>/', Product.as_View(), name="productdetail"),
Views.py
class ArticleDetail(DetailView):
def get_object(self):
slug = self.kwargs.get('slug')
article = get_object_or_404(Article.objects.published(), slug=slug)
ip_address = self.request.user.ip_address
if ip_address not in article.hits.all():
article.hits.add(ip_address)
return article
class CategoryList(ListView):
paginate_by = 5
template_name = 'blog/category_list.html'
def get_queryset(self):
global category
slug = self.kwargs.get('slug')
category = get_object_or_404(Category.objects.active(), slug=slug)
return category.articles.published()
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['category'] = category
return context
urls.py
urlpatterns = [
path('<slug:slug>', ArticleDetail.as_view(), name="detail"),
path('<slug:slug>', CategoryList.as_view(), name="category"),
]
This is my django blog codes, I don't want write article or category & ... in urls, just slug
mysite .com/article-slug ... mysite .com/category-slug
now one url pattern ok but another url pattern don't work and have 404 error :(
@WillemVanOlsem is right, you will have to write a view like this:
from django.http import HttpResponseNotFound
def slug_router(request, slug):
if Category.objects.filter(slug=slug).exists():
return CategoryList.as_view()(request, slug=slug)
elif Article.objects.filter(slug=slug).exists():
return ArticleDetail.as_view()(request, slug=slug)
else:
return HttpResponseNotFound('404 Page not found')
And then
urlpatterns = [
path('<slug:slug>', slug_router, name="slug"),
]
... if I'm not mistaken. This should be the jist of it. I didn't test this code, just typed it in here, so let me know if it doesn't work, I'll help to fix it.
Note that you'll have a preference if there are Articles with the same slug as some Categories.