TypeError：在include（）的情况下，view必须是可调用的或list / tuple

Question

我是django和python的新手。在URL映射到视图期间，我收到以下错误： TypeError：在include（）。

的情况下，view必须是可调用的或list / tuple

网址。 py代码： -

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


urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^posts/$', "posts.views.post_home"), #posts is module and post_home 
]                                              # is a function in view. 

views.py代码： -

from django.shortcuts import render
from django.http import HttpResponse
# Create your views here.
#function based views

def post_home(request):
    response = "<h1>Success</h1>"
    return HttpResponse(response)

回溯

Answer 1

在1.10中，您无法再将导入路径传递给url()，您需要传递实际的视图功能：

from posts.views import post_home

urlpatterns = [
    ...
    url(r'^posts/$', post_home),
]        

Answer 2

用此

替换您的管理网址模式

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

所以你的urls.py成为：

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


urlpatterns = [
    url(r'^admin/', include(admin.site.urls)),
    url(r'^posts/$', "posts.views.post_home"), #posts is module and post_home 
] 

管理员网址可以通过include（1.9之前）调用。

Answer 3

对于Django 1.11.2
在主urls.py中写道：

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

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

在appname / urls.py文件中写：

from django.conf.urls import url
from . import views

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

Answer 4

答案在 project-dir / urls.py

中

Including another URLconf
    1. Import the include() function: from django.conf.urls import url, include
    2. Add a URL to urlpatterns:  url(r'^blog/', include('blog.urls'))

Answer 5

为了补充@knbk的答案，我们可以使用以下模板：

，如1.9：

from django.conf.urls import url, include

urlpatterns = [
    url(r'^admin/', admin.site.urls), #it's not allowed to use the include() in the admin.urls
    url(r'^posts/$', include(posts.views.post_home), 
] 

应该在1.10中：

from your_project_django.your_app_django.view import name_of_your_view

urlpatterns = [
    ...
    url(r'^name_of_the_view/$', name_of_the_view),
]

请记住在your_app_django＆gt;＆gt;中创建views.py用于渲染视图的函数。

Answer 6

您需要传递实际的视图功能

来自posts.views导入post_home

urlpatterns = [     ...     url（r＆＃39; ^ posts / $＆＃39;，post_home）， ]

这很好用！ 您可以在URL Dispatcher Django阅读 在这里Common Reguler Expressions Django URLs

Answer 7

您将需要导入post_home：

from django.contrib import admin
from django.urls import path,include
from django.conf.urls import url
from posts.views import post_home

urlpatterns = [
    path('admin/', admin.site.urls),
    path('posts/',post_home),
]