Django：如何通过给出命名空间来获取模板的url？(Django: How can i get the url of a template by giving the namespace?)

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，而不是这样：

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

- 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'})

- 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'})

Thanks in advance.

最满意答案

模板名称在视图中是硬编码的。 你还可以做的是你可以从url模式传递模板名称，有关详细信息，请参阅此处 ：

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']

Template names are hard coded within the view. What you can also do is that you can pass the template name from the url pattern, for more details see here:

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

Then in view get the template name:

def index(request, **kwargs): template_name = kwargs['template_name']Django：如何通过给出命名空间来获取模板的url？(Django: How can i get the url of a template by giving the namespace?)

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，而不是这样：

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

- 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'})

- 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'})

Thanks in advance.

最满意答案

模板名称在视图中是硬编码的。 你还可以做的是你可以从url模式传递模板名称，有关详细信息，请参阅此处 ：

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']

Template names are hard coded within the view. What you can also do is that you can pass the template name from the url pattern, for more details see here:

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

Then in view get the template name:

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