Django Oscar改变了URL模式

Question

我已经设置了django-oscar项目，我正在尝试配置URL。我的目标是将/catalogue改为/catalog。

根据文档，我在app.py中添加了myproject/app.py

MyProject的/ app.py

from django.conf.urls import url, include
from oscar import app


class MyShop(app.Shop):
    # Override get_urls method
    def get_urls(self):
        urlpatterns = [
            url(r'^catalog/', include(self.catalogue_app.urls)),
            # all the remaining URLs, removed for simplicity
            # ...
        ]
        return urlpatterns


application = MyShop()

MyProject的/ urls.py

from django.conf.urls import url, include
from django.contrib import admin
from . import views
from .app import application

urlpatterns = [
    url(r'^i18n/', include('django.conf.urls.i18n')),

    url(r'^admin/', admin.site.urls),

    url(r'', application.urls),

    url(r'^index/$',views.index, name = 'index'),
]

项目服务器运行没有任何错误，但当我尝试localhost:8000/catalog我得到

/ catalog /'customer'中的NoReverseMatch不是注册的命名空间。

预期的输出是localhost:8000/catalog应该返回目录页面。

Answer 1

扩展c.grey's answer以指定如何替换而不是添加网址 -

from django.conf.urls import url, include
from oscar import app


class MyShop(app.Shop):
    def get_urls(self):
        urls = super(MyShop, self).get_urls()
        for index, u in enumerate(urls):
            if u.regex.pattern == r'^catalogue/':
                urls[index] = url(r'^catalog/', include(self.catalogue_app.urls))
                break
        return urls


application = MyShop()

Answer 2

你可以试试这个

在app.py中

from django.conf.urls import url, include
from oscar import app

class MyShop(app.Shop):
    # Override get_urls method
    def get_urls(self):
        urls = [
            url(r'^catalog/', include(self.catalogue_app.urls)),
            # all the remaining URLs, removed for simplicity
            # ...
        ]
        urls = urls + super(MyShop,self).get_urls()
        return urls


application = MyShop()

在你的urls.py中你可以简单地添加它

from myproject.app import application as shop

 url(r'', shop.urls),

希望对你有所帮助

Answer 3

您需要包含URL，而不是直接引用它们。

url(r'', include('application.urls')),