我目前在urls.py中有一个条目,它为我的bug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>[^/]+)/$',
'tagging.views.tagged_object_list',
{
'queryset_or_model': Bug,
'template_name' : 'tag/lone.html'}),
# Uncomment the next line to enable the admin:
(r'^admin/', include(admin.site.urls)),
)因此,如果我指定一个url,比如说bugs/tagged/firefox,它会显示出火狐标签。如何让它通过多个标签过滤出来?例如:firefox+css将返回所有带有firefox和css标签的对象。
发布于 2010-06-21 22:16:56
您必须构建自己的视图,而不是使用tagging.views.tagged_object_list。
(r'^bugs/tagged/(?P<tags>[-\w+]+)/$', your_tag_view)在您的视图中,获取要搜索的标签的列表:
tags = tags.split('+')然后,使用TaggedItem.objects.get_by_model查询,它可以方便地接受标记列表:
from tagging.models import TaggedItem
bugs = TaggedItem.objects.get_by_model(Bug, tags)https://stackoverflow.com/questions/3077385
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