當我嘗試使用Django運行服務器時獲取TypeError？

Question

我發現接近我問的問題，但我在那裏找到的解決方案對我沒有用。使用Django版本1.10.1

輸出到

蟒蛇manage.py runserver命令

Performing system checks... 

Unhandled exception in thread started by <function wrapper at 0xb6838064> 
Traceback (most recent call last): 
    File "/usr/local/lib/python2.7/dist-packages/django/utils/autoreload.py", line 226, in wrapper 
    fn(*args, **kwargs) 
    File "/usr/local/lib/python2.7/dist-packages/django/core/management/commands/runserver.py", line 121, in inner_run 
    self.check(display_num_errors=True) 
    File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 374, in check 
    include_deployment_checks=include_deployment_checks, 
    File "/usr/local/lib/python2.7/dist-packages/django/core/management/base.py", line 361, in _run_checks 
    return checks.run_checks(**kwargs) 
    File "/usr/local/lib/python2.7/dist-packages/django/core/checks/registry.py", line 81, in run_checks 
    new_errors = check(app_configs=app_configs) 
    File "/usr/local/lib/python2.7/dist-packages/django/core/checks/urls.py", line 14, in check_url_config 
    return check_resolver(resolver) 
    File "/usr/local/lib/python2.7/dist-packages/django/core/checks/urls.py", line 24, in check_resolver 
    for pattern in resolver.url_patterns: 
    File "/usr/local/lib/python2.7/dist-packages/django/utils/functional.py", line 35, in __get__ 
    res = instance.__dict__[self.name] = self.func(instance) 
    File "/usr/local/lib/python2.7/dist-packages/django/urls/resolvers.py", line 313, in url_patterns 
    patterns = getattr(self.urlconf_module, "urlpatterns", self.urlconf_module) 
    File "/usr/local/lib/python2.7/dist-packages/django/utils/functional.py", line 35, in __get__ 
    res = instance.__dict__[self.name] = self.func(instance) 
    File "/usr/local/lib/python2.7/dist-packages/django/urls/resolvers.py", line 306, in urlconf_module 
    return import_module(self.urlconf_name) 
    File "/usr/lib/python2.7/importlib/__init__.py", line 37, in import_module 
    __import__(name) 
    File "/home/sidgupta234/Desktop/Hack/guide/guide/urls.py", line 21, in <module> 
    url(r'^$', "hacko.views.home", name='home'), 
    File "/usr/local/lib/python2.7/dist-packages/django/conf/urls/__init__.py", line 85, in url 
    raise TypeError('view must be a callable or a list/tuple in the case of include().') 
TypeError: view must be a callable or a list/tuple in the case of include(). 

urls.py文件

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

urlpatterns = [ 
    url(r'^admin/', admin.site.urls), 
    url(r'^$', "hacko.views.home", name='home'), 
    url(r'^index/$', "hacko.views.index", name='index'), 
] 

views.py文件

from django.shortcuts import render 
import requests 
from django.http import HttpResponse 
import unicodedata 
# Create your views here. 
def home(request): 

    return render(request,"sample.html"); 

def index(request): 
    print "i was called" 
    if(request.GET.get('q', '')): 
     theUrl="http://en.wikipedia.org/w/api.php?action=query&prop=extracts&format=json&titles="+request.GET.get('q', '') 
     r = requests.get(theUrl, auth=('user', 'pass')); 
     print r.text 
     return HttpResponse(r.text, content_type='hacko/json') 
    elif(request.GET.get('p', '')): 
     theUrl="https://en.wikipedia.org/w/api.php?action=query&prop=extracts&format=json&exintro=&titles="+request.GET.get('p', '') 
     r = requests.get(theUrl, auth=('user', 'pass')); 
     print r.text 
     return HttpResponse(r.text, content_type='hacko/json') 
    else: 
     key="f7ee811c-3f3d-47b4-ad77-b31fe1831b2f" 
     r="http://www.dictionaryapi.com/api/v1/references/thesaurus/xml/"+request.GET.get('r', '')+"?key="+key 
     r=requests.get(r); 
     r=r.text 
     r=unicodedata.normalize('NFKD', r).encode('ascii','ignore') 
     return HttpResponse(r,content_type="text/plain") 

Answer 1

更新您的urls.py文件像下面

from django.conf.urls import url 
from django.contrib import admin 
from hacko import views #importing views from app 

urlpatterns = [ 
    url(r'^admin/', admin.site.urls), 
    url(r'^$', views.home, name='home'), 
    url(r'^index/$', views.index, name='index'), 
] 

爲您的reference。還請按照與您的django version匹配的正確文章或教程進行操作。它總是很好地遵循Django官方網站。

Answer 2

在1.10，可以不再通過進口路徑URL（），您需要通過實際的視圖功能：

from django.conf.urls import url 
from django.contrib import admin 
from hacko import views 
urlpatterns = [ 
    url(r'^admin/', admin.site.urls), 
    url(r'^$', views.home, name='home'), 
    url(r'^index/$', views.index, name='index'), 
]