在 Django 上找不到"empty slug"模式的页面

Question

在我的应用程序中，我想显示一个基于特定url的项目列表。例如，当我键入mysite.com/restaurants/chinese/时，我想显示所有的中餐馆。如果我键入mysite.com/restaurants/american/，我想显示所有的美国餐厅等等。但当我键入mysite.com/restaurants/时，我想展示所有的餐厅，所以我写了这个代码：

urls.py

from django.conf.urls import url, include
from django.contrib import admin

urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^restaurants/', include('restaurants.urls')),
]

餐馆/网址.py

from django.conf.urls import url
from django.views.generic import ListView
from .views import RestaurantLocationListView
urlpatterns = [
url(r'^(?P<slug>w+)/$', RestaurantLocationListView.as_view()),
]

餐馆/景观.py

from django.db.models import Q
from django.shortcuts import render
from django.views.generic import ListView
from .models import RestaurantLocation
class RestaurantLocationListView(ListView):
def get_queryset(self):
slug = self.kwargs.get('slug')
if slug:
queryset = RestaurantLocation.objects.filter(
Q(category__iexact=slug) | Q(category__icontains=slug)
)
else:
queryset = RestaurantLocation.objects.all()
return queryset

一切都很好，除了我只放mysite.com/restaurants/。这给了我一个404错误，而不是所有餐馆的列表，我不知道为什么。你们能帮我吗？

Answer 1

似乎是一个url问题。在你的restaunts/url.py中，你需要为它添加一个url。

urlpatterns = [
url(r'^$', RestaurantLocationListView.as_view()), # add it before
url(r'^(?P<slug>w+)/$', RestaurantLocationListView.as_view()),
]