[英]django url from a django app is giving 404 error
demo > url.py (This is main django project url file)演示 > url.py (这是主要的 django 项目 url 文件)
from django.contrib import admin
from django.urls import path, include
from . import views
urlpatterns = [
path('admin/', admin.site.urls),
path('', views.home, name='home'),
path('about', views.about, name='about'),
path('contact', views.contact, name='contact'),
path(r'^user/', include('user.urls')),
]
user > url.py (this is a user app inside django project's url file)用户 > url.py(这是 django 项目的 url 文件中的用户应用程序)
from django.urls import path
from . import views
urlpatterns = [
path(r'^/login', views.login, name='login'),
]
now i am trying to open http://127.0.0.1:8000/user/login but its giving me 404 error.现在我正在尝试打开http://127.0.0.1:8000/user/login但它给了我 404 错误。
You don't use a regex for path(…)
[Django-doc] , for regexes, you usere_path(…)
[Django-doc] .对于path(…)
[Django-doc] ,您不使用正则表达式,对于正则表达式,您使用re_path(…)
[Django-doc] 。 But here it is not necessary: you can define this with path(…)
:但在这里没有必要:您可以使用path(…)
定义它:
from django.contrib import admin
from django.urls import path, include
from . import views
urlpatterns = [
path('admin/', admin.site.urls),
path('', views.home, name='home'),
path('about', views.about, name='about'),
path('contact', views.contact, name='contact'),
path('user/', include('user.urls')),
]
and for user/urls.py
:对于user/urls.py
:
urlpatterns = [
path('login/', views.login, name='login'),
]
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