TypeError:在include()的情况下，视图必须是可调用的或列表/元组

I am new to django and python. During url mapping to views i am getting following error: TypeError: view must be a callable or a list/tuple in the case of include().

我是django和python的新手。在到视图的url映射过程中，我得到了以下错误:TypeError: view必须是可调用的，如果是include()，则必须是列表/元组。

Urls. py code:-

url。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 code:-

的观点。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)

Traceback

回溯

6 个解决方案

#1

In 1.10, you can no longer pass import paths to url(), you need to pass the actual view function:

在1.10中，不能再将导入路径传递给url()，需要传递实际的视图函数:

from posts.views import post_home

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

#2

Replace your admin url pattern with this

用这个替换您的管理url模式

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

So your urls.py becomes :

所以你的url。py就变成:

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 
]

admin urls are callable by include (before 1.9).

admin url可以通过include调用(在1.9之前)。

#3

For Django 1.11.2
In the main urls.py write :

对于主要url中的Django 1.11.2。py写:

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")),
]

And in the appname/urls.py file write:

和浏览器名称/ url。py文件编写:

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

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

#4

Answer is in project-dir/urls.py

答案是project-dir / urls . py

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'))

#5

Just to complement the answer from @knbk, we could use the template below:

为了补充@knbk的答案，我们可以使用下面的模板:

as is in 1.9:

是在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), 
]

as should be in 1.10:

如1.10所示:

from your_project_django.your_app_django.view import name_of_your_view

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

Remember to create in your_app_django >> views.py the function to render your view.

请记住在your_app_django >>视图中创建。函数的作用是:渲染视图。

#6

You need to pass actual view function

您需要传递实际的视图函数

from posts.views import post_home

从职位。视图导入post_home

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

urlpattern =[…url(r / $ ^帖子,post_home),)

This works fine! You can have a read at URL Dispatcher Django and here Common Reguler Expressions Django URLs

这个没问题!您可以在URL分派器Django和这里常见的正则表达式Django URL上进行读取

#1