Django URL不接受参数

Question

我真的很难理解如何将参数和keword参数发送到Django url。以下是案例研究：

我有一个使用通用基本视图的视图：

class CartView(View):
   def get(self, request, *args, **kwargs):
       item = request.GET.get('item')
       qty = request.GET.get('qty')
       print item, qty
       return HttpResponseRedirect('/')

有了上面的视图，我可以使用url中的aurguments，比如"localhost:8000/cart/?item=4&qty=200"，并在terminal中打印带有数量的条目。你知道吗

一旦我对代码做了如下更改：

from carts.models import Cart, CartItem
from products.models import Variation


class CartView(View):
    def get(self, request, *args, **kwargs):
        item_id = request.GET.get('item')
        if item_id:
            item_instance = get_object_or_404(Variation, id=item_id)
            qty = request.GET.get('qty')
            cart = Cart.objects.all()[0]
            cart_item = CartItem.objects.get_or_create(cart=cart, item=item_instance)[0]
            cart_item.quantity = qty
            cart_item.save()
            print cart_item
        return HttpResponseRedirect('/')

用同样的方法传递参数，比如"localhost:8000/cart/?item=4&qty=200"，它向我显示了错误：

404 Page Not Found No Variation matches the given query.

你知道吗网址.py你知道吗

urlpatterns = [
    url(r'^home/$', 'newsletter.views.home', name='home'),
    url(r'^contact/$', 'newsletter.views.contact', name='contact'),
    url(r'^about/$', 'project.views.about', name='about'),
    url(r'^admin/', include(admin.site.urls)),
    url(r'^accounts/', include('registration.backends.default.urls')),
    url(r'^cart/$', CartView.as_view(), name='cart'),
    url(r'^', include('products.urls')),
    url(r'^categories/', include('products.urls_categories')),

Answer 1

404 Page Not Found

No Variation matches the given query.

此信息来自您的电话：

item_instance = get_object_or_404(Variation, id=item_id)

这意味着没有与给定的id匹配的Variation对象。你知道吗