python - 从所选标签返回对象列表

Question

我想找出我认为很容易找到答案的问题。我在一个项目中使用 django-taggit，我只是想在选择标签时返回一个对象列表。我试过这个:

How do I create list and detail views for django-taggit?

但我无法让它工作。它只是呈现一个空白页面。我认为问题出在我的模板代码中。也许有人可以指出我的方向。任何帮助将不胜感激..非常感谢。

这是我的代码:

模型.py

from taggit.managers import TaggableManager
from django.template.defaultfilters import slugify
from ckeditor.fields import RichTextField 
from taggit.models import TaggedItemBase

class Tagged(TaggedItemBase):
content_object = models.ForeignKey('Shows')

class Shows(models.Model):

title = models.CharField(max_length=40)
slug = models.SlugField(null=True, blank=True, unique=True)
tags = TaggableManager(through=Tagged)
hosts = models.ManyToManyField('Host', blank=True, null=True)
featured = models.BooleanField(default=False)
thumbnail = FilerImageField(related_name="thumbnail", help_text="Image should be: 550 X 350.")
playing_next = models.DateTimeField(null=True, blank=True)
description = RichTextField()

views.py:

class TaggedList(ListView):
queryset = Shows.objects.all()
paginate_by = 10
template_name = "tagged.html"

def get_queryset(self):
    return Shows.objects.filter(tags__name__in=[self.kwargs['tag']])

urls.py:

urlpatterns = patterns('radio.views',
                   url(r'^$', 'main', name='app_main'),
                   url(r'^(?P<slug>[^\.]+)/detail/$', 'detail_view', name='detailsview'),
                   url(r'^(?P<tag>\w+)/$', TaggedList.as_view()),
                   url(r'^tagged/(?P<tag>\w+)/$', TaggedList.as_view())

                   )

Answer 1

模板代码:

{% for objects in object_list %}
    {{ objects.title }}
    {{ objects.tag }}
    {{ objects.slug }} 
     ------
{% endfor %}