python - Pass slug field to url -

i'm trying create detailed view of model: product.

but, i'm not getting how pass slug-field url.

this detailed view:

def single_product(request):     product = get_object_or_404(prods, slug=slug)     return render_to_response('teste.html', locals(), context_instance=requestcontext(request)) 

this model:

class product(models.model):     name = models.charfield(max_length=500)     #inserir slugify na url produto     slug = models.slugfield(max_length=500)     category = models.foreignkey(category)     image = models.imagefield(upload_to='thumbs/')     created = models.datetimefield(auto_now=true, auto_now_add=false)     updated = models.datetimefield(auto_now=true, auto_now_add=true) 

and value of slug product 1:'kinect-xbox-360' when try run: [localhost]/kinect-xbox-360/ message:

typeerror @ /kinect-xbox-360/ 'str' object not callable 

you need add slug appropriate url in urls.py. passed parameter in view:

urls.py

... url(r'^product/(?p<slug>[\w-]+)/$', 'views.single_product', name='detail'), ... 

read more url dispatcher

views.py

def single_product(request, slug):     product = get_object_or_404(prods, slug=slug)     return render_to_response('test.html', locals(), context_instance=requestcontext(request)) 

read more writing function-based views