django - 类型错误 : view must be a callable or a list/tuple in the case of include()

Question

我是 Django 和 python 的新手。在 url 映射到 View 期间，我收到以下错误:
类型错误:在 include() 的情况下， View 必须是可调用的或列表/元组。

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)

追溯

Answer 1

在 1.10 中，您不能再将导入路径传递给 url() ，您需要传递实际的 View 函数:

from posts.views import post_home

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