1

我目前有一个条目，urls.py其中为我的错误获取单独的永久链接：

from django.conf.urls.defaults import *

from tagging.views import tagged_object_list

from bugs.models import Bug

# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
    # Example:
    # (r'^workarounds/', include('workarounds.foo.urls')),

    # Uncomment the admin/doc line below and add 'django.contrib.admindocs' 
    # to INSTALLED_APPS to enable admin documentation:
    # (r'^admin/doc/', include('django.contrib.admindocs.urls')),

    (r'^$', 'django.views.generic.simple.direct_to_template', {'template':'homepage.html'}),

    (r'^bugs/(?P<slug>[-\w]+)/$', 'bugs.views.bug_detail'),
    (r'^bugs/tagged/(?P<tag>[^/]+)/$', 
    'tagging.views.tagged_object_list',
    {
        'queryset_or_model': Bug,
        'template_name' : 'tag/lone.html'}),
    # Uncomment the next line to enable the admin:
    (r'^admin/', include(admin.site.urls)),
)

因此，如果我指定一个 url，bugs/tagged/firefox它会调出 firefox 标签。我怎样才能让它被多个标签过滤掉？例如：将返回所有用andfirefox+css标记的对象。firefoxcss

4

1 回答 1

3

您必须构建自己的视图，而不是使用tagging.views.tagged_object_list.

(r'^bugs/tagged/(?P<tags>[-\w+]+)/$', your_tag_view)

在您看来，获取您正在搜索的标签的列表：

tags = tags.split('+')

然后，使用TaggedItem.objects.get_by_model查询，它会方便地接受标签列表：

from tagging.models import TaggedItem
bugs = TaggedItem.objects.get_by_model(Bug, tags)
于 2010-06-21T14:16:55.597 回答