将slug字段传递给url

Question

我正在尝试创建我的模型的详细视图：Product。

但是，我没有得到如何将slug-field传递到url。

这是我的详细视图：

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

这是我的模特：

class Product(models.Model):
    name = models.CharField(max_length=500)
    #inserir slugify na url do 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)

这就是产品1的slug价值：'kinect-xbox-360' 但是当我尝试运行时：[localhost] / kinect-xbox-360 / 我收到这条消息：

TypeError at /kinect-xbox-360/
'str' object is not callable

Answer 1

您需要将slug添加到urls.py中的相应网址。然后，它将作为视图中的参数传递：

<强> urls.py

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

read more about the url dispatcher

<强> 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 about writing function-based views