django下的报错

tz104

django 1.5
python 2.6.5
模仿着django step by step:http://www.lhelper.org/dev/djang ... test/doc/tut07.html 的例子学习
在例子“创建通讯录”中，遇到了报错，如下：
TypeError at /address/
__init__() takes exactly 1 non-keyword argument (2 given)
Request Method:        GET
Request URL:        http://192.168.1.111:8080/address/
Django Version:        1.5.1
Exception Type:        TypeError
Exception Value:       
__init__() takes exactly 1 non-keyword argument (2 given)
Exception Location:        /usr/lib/python2.6/site-packages/django/core/handlers/base.py in get_response, line 115
Python Executable:        /usr/bin/python
Python Version:        2.6.5
Python Path:       
['/opt/Django_Pro',
'/opt/Django_Pro/Django_Pro',
'/opt/Django_Pro',
'/usr/lib/python2.6/site-packages/distribute-0.6.49-py2.6.egg',
'/usr/lib/python26.zip',
'/usr/lib/python2.6',
'/usr/lib/python2.6/plat-linux2',
'/usr/lib/python2.6/lib-tk',
'/usr/lib/python2.6/lib-old',
'/usr/lib/python2.6/lib-dynload',
'/usr/lib/python2.6/site-packages',
'/usr/lib/python2.6/site-packages/gtk-2.0'
不知如何破？
address下的urs.py如下：
from django.conf.urls import *
from django.views.generic.list import ListView
from address.models import Address

info_dict = {
    'queryset': Address.objects.all(),
}

urlpatterns = patterns('',
   url(r'^/?$','django.views.generic.list.ListView',info_dict),
)