Python 简单留言簿django:\uuuu init\uuuu（）接受1个位置参数，但给出了2个

我是Django的新手，尝试制作一个简单的留言簿应用程序来适应环境。我犯了以下错误，但找不到错误：

异常值：\u init\u（）接受1个位置参数，但给出了2个。

from django.db import models
from django.contrib.auth.models import User
from django.contrib import admin

class Bericht(models.Model):
    titel = models.CharField(max_length=50)
    auteur = models.ForeignKey(User, blank=True)
    email = models.EmailField(max_length=75)
    inhoud = models.TextField(max_length=10000, blank=True)
    datum = models.DateTimeField(auto_now_add=True)

    def __str__(self):
        return str(self.auteur) + " : " + str(self.titel)

    class Meta:
        verbose_name_plural = "berichten"

class BerichtAdmin(admin.ModelAdmin):
    list_display = ["auteur", "datum", "titel"]
    list_filter = ["datum", "auteur"]

admin.site.register(Bericht, BerichtAdmin)

景色

from django.shortcuts import render
from django.views.generic import ListView
from Gastenboek.models import *

class BerichtListView(ListView):
    model = Bericht.objects.all()
    template_name = 'template/bericht_lijst.html'
    paginate_by = 10
    context_object_name = "bericht_lijst"
# Create your views here.

url.py

from django.conf.urls import patterns, include, url

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    # Examples:
    # url(r'^$', 'Niels.views.home', name='home'),
    # url(r'^blog/', include('blog.urls')),

    url(r'^admin/', include(admin.site.urls)),
    (r"^(\d+)/$", 'Gastenboek.views.BerichtListView'),
    (r"", 'Gastenboek.views.BerichtListView'),
)

回溯

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/

Django Version: 1.6.1
Python Version: 3.3.3
Installed Applications:
('django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'Gastenboek')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware')


Traceback:
File "C:\Python33\lib\site-packages\django\core\handlers\base.py" in get_response
  114.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)

Exception Type: TypeError at /
Exception Value: __init__() takes 1 positional argument but 2 were given

在您的URL.py中：

您丢失了。as_view（）

将其更改为：

(r"^(\d+)/$", Gastenboek.views.BerichtListView.as_view()),
(r"", Gastenboek.views.BerichtListView.as_view()),

---

将完整的错误消息粘贴到这里。我认为您的问题来自model=Bericht.objects.all（），它应该是model=Bericht或您的URL未添加。正如_view（）所示，您在此处显示的任何代码中都没有问题。您应该始终粘贴完整的回溯。不过我怀疑这个问题在你的URL.py中。看看：这个问题可能非常相似。谢谢大家，我根据要求更新了一些信息。现在我收到一个错误，说Gastenboek没有定义。当我这样做时：从Gastenboek导入*，它说没有名为site的模块。你不应该使用通配符。。。尝试导入Gastenboek.views。BerichtListView@Alvaro，你帮我省去了很多麻烦，谢谢@GED125很高兴这仍然有用：）