使用装饰器login_required的Django通用视图

Question

我是django的新手,我在djangoproject.com上完成了4部分教程

我的问题是我想在我的民意调查应用程序上进行登录身份验证.我使用装饰器@login_required并且它正常工作但在我的views.py下我只有vote()方法.

我的views.py在"polls文件夹"下

from django.shortcuts import render_to_response, get_object_or_404
from django.http import HttpResponseRedirect, HttpResponse
from django.contrib.auth.decorators import login_required
from django.views.decorators.cache import never_cache
from django.core.urlresolvers import reverse
from django.template import RequestContext
from polls.models import Poll, Choice


@login_required
@never_cache
def vote(request, poll_id):
    p = get_object_or_404(Poll, pk=poll_id)
    try:
        selected_choice = p.choice_set.get(pk=request.POST['choice'])
    except (KeyError, Choice.DoesNotExist):
        return render_to_response('polls/detail.html', {
            'poll': p,
            'error_message': "You didn't select a choice.",
        }, context_instance=RequestContext(request))
    else:
        selected_choice.votes += 1
        selected_choice.save()
        return HttpResponseRedirect(reverse('poll_results', args=(p.id,)))
    return HttpResponse("You're voting on poll %s." % poll_id) 

我的urls.py在"polls文件夹"下

from django.conf.urls.defaults import patterns, include, url
from django.views.generic import DetailView, ListView
from polls.models import Poll

urlpatterns = patterns('',
    url(r'^$',
        ListView.as_view(
            queryset = Poll.objects.order_by('-pub_date')[:5],
            context_object_name = 'latest_poll_list',
            template_name = 'polls/index.html'), name='poll_lists'),
    url(r'^(?P<pk>\d+)/$',
        DetailView.as_view(
            model = Poll,
            template_name = 'polls/detail.html'), name='poll_details'),
    url(r'^(?P<pk>\d+)/results/$',
        DetailView.as_view(
            model = Poll,
            template_name = 'polls/results.html'), name = 'poll_results'),
    url(r'^(?P<poll_id>\d+)/vote/$', 'polls.views.vote'),
)

在我的urls.py下我使用了通用视图.

问题是:如何在民意调查应用程序的"索引"下放置登录.这样用户在他/她可以查看可用的民意调查之前将首先登录.

当前是:我在views.py和方法vote()下使用了登录,在投票后需要登录.

有人可以帮帮我吗？

谢谢,贾斯汀

Answer 1

第一种方法

在urls.py中:

urlpatterns = patterns('',
    url(r'^$',
        login_required(ListView.as_view(
            queryset = Poll.objects.order_by('-pub_date')[:5],
            context_object_name = 'latest_poll_list',
            template_name = 'polls/index.html'), name='poll_lists')),
)

第二种方法

在views.py中:

class IndexView(ListView):
    queryset = Poll.objects.order_by('-pub_date')[:5]
    context_object_name = 'latest_poll_list'
    template_name = 'polls/index.html'

    @method_decorator(login_required)
    def dispatch(self, request, *args, **kwargs):        
        return super(IndexView, self).dispatch(request, *args, **kwargs)

然后在urls.py中

urlpatterns = patterns('',
        url(r'^$',
            IndexView.as_view(), name='poll_lists'),
    )

Answer 2

只是提供一个可能更新的答案，

我会将它们移动到视图文件并使用LoginRequiredMixin，

from django.views.generic import (
    ListView,
    DetailView
)
from django.contrib.auth.mixins import LoginRequiredMixin


class PollsListView(LoginRequiredMixin, ListView):
    model = Poll
    template_name = 'polls/index.html'