Django将所有未捕获的URL路由到包含的urls.py [英] Django route all non-catched urls to included urls.py

问题描述

我希望每个不以"api"开头的网址都使用foo/urls.py

I want every url that does not start with 'api' to use the foo/urls.py

urls.py

from django.conf.urls import include, url
from foo import urls as foo_urls

urlpatterns = [
url(r'^api/', include('api.urls', namespace='api')),
url(r'^.*/$', include(foo_urls)),
]    

foo/urls.py

foo/urls.py

from django.conf.urls import include, url
from foo import views

urlpatterns = [
url(r'a$', views.a),
]    

这行不通，知道吗?

推荐答案

如果您想捕获所有网址格式，请使用:

If you want a catch all url pattern, use:

url(r'^', include(foo_urls)),

来自文档:

每当Django遇到 include()时，它都会截断直到该点匹配的URL的任何部分，并将剩余的字符串发送到包含的URLconf中以进行进一步处理.

Whenever Django encounters include() it chops off whatever part of the URL matched up to that point and sends the remaining string to the included URLconf for further processing.

在您当前的代码中，正则表达式 ^.*/$ 与整个URL /a/匹配.这意味着没有多余的东西可以传递给 foo_urls .

In your current code, the regex ^.*/$ matches the entire url /a/. This means that there is nothing remaining to pass to foo_urls.

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