Django：如何通过给出命名空间来获取模板的url？

Question

当我渲染模板时，我想通过给出命名空间值而不是路径来检索模板的URL。例如，而不是：

return render(request, 'base/index.html', {'user':name})

我希望能够做到以下几点：

from django.shortcuts import render
from django.core.urlresolvers import reverse

return render(request, reverse('base:index'), {'user':name})

但上面会产生错误。我该怎么做？有没有办法将命名空间赋予函数并获取实际路径？

扩展示例：

- urls.py

from django.conf.urls.defaults import patterns, include, url

urlpatterns = patterns('',
    url(r'^', include('base.urls', namespace='base')),
)

- app base：urls.py

from django.conf.urls.defaults import patterns, url

urlpatterns = patterns('base.views',
   url(r'^/?$', 'index', name='index'),
)

- app base：views.py

from django.shortcuts import render
from django.core.urlresolvers import reverse

def homepage(request):
   '''
   Here instead of 'base_templates/index.html' i would like to pass
   something that can give me the same path but by giving the namespace
   '''
   return render(request, 'base_templates/index.html', {'username':'a_name'})

提前致谢。

Answer 1

模板名称在视图中是硬编码的。您还可以做的是，您可以从网址格式传递模板名称，有关详细信息，请参阅here：

from django.conf.urls.defaults import patterns, url

    urlpatterns = patterns('base.views',
       url(r'^/?$', 'index',
          {'template_name': 'base_templates/index.html'},
          name='index'),
       )

然后在视图中获取模板名称：

def index(request, **kwargs):
    template_name = kwargs['template_name']