2013-06-20 71 views
0

我在这里继例如:http://django-tastypie.readthedocs.org/en/latest/tutorial.html 我的urls.py:问题测试的Django tastypie

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

from django.conf.urls.defaults import * 
from ristoturisto.api import EntryResource 

entry_resource = EntryResource() 
admin.autodiscover() 
urlpatterns = patterns('', 
    url(r'^admin/', include(admin.site.urls)), 
    (r'^blog/', include('ristoturisto.urls')), #this basically points to it self? 
    (r'^api/', include(EntryResource.urls)), 

api.py

from tastypie.resources import ModelResource 
from locations.models import tours 


class EntryResource(ModelResource): 
    class Meta: 
     queryset = tours.objects.all() 
     resource_name = 'Tours' 

型号:

class tours(models.Model): 
    name = models.CharField(max_length=255) 
    categories = models.ForeignKey('categories') 
    icon = models.CharField(max_length=255) 
    publishdate = models.CharField(max_length=255) 
    locations = models.ManyToManyField('geolocations') 

我得到的错误是:

ImproperlyConfigured在/ API /旅游

当我试图访问:http://127.0.0.1:8000/api/tours?format=json

哪里Entity_resource得到它的URL从公司?这不是在这个例子中?

回答

1

您可以使用类而不是这个类entry_resource实例EntryResource

(r'^api/', include(EntryResource.urls)), 

改变它:

(r'^api/', include(entry_resource.urls)), 
+0

老天爷你是对的!我现在有一个新的错误,但是,谢谢你,为期两天......你的先生是一个真正的绅士! – R0b0tn1k

+1

你使用任何代码检查器?我使用flake8,并在这种情况下通知我(例如,变量未使用)。 https://pypi.python.org/pypi/flake8 –