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,我遇到以下错误: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)

追溯

Answer 1

在1.10中,您无法再传递导入路径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),
]