我希望每个不以“api”开头的 url 都使用 foo/urls.py
网址.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'^.*/$', include(foo_urls)),
]
foo/urls.py
from django.conf.urls import include, url
from foo import views
urlpatterns = [
url(r'a$', views.a),
]
这行不通，知道吗？
最佳答案
如果你想要捕获所有 url 模式，请使用:
url(r'^', include(foo_urls)),
来自 the docs :
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 ^.*/$ 匹配整个 url /a/。这意味着没有任何东西可以传递给 foo_urls。
关于python - Django 将所有未捕获的 url 路由到包含的 urls.py，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/31248126/