Question

为什么这不起作用

handler500 = TemplateView.as_view(template_name="500.html")

我得到以下异常：

Traceback (most recent call last):
  File "/usr/lib/python2.6/wsgiref/handlers.py", line 94, in run    
    self.finish_response()
  File "/usr/lib/python2.6/wsgiref/handlers.py", line 134, in finish_response
    for data in self.result:
  File "/home/hatem/projects/leadsift_app/.virtualenv/lib/python2.6/site-packages/django/template/response.py", line 117, in __iter__
    raise ContentNotRenderedError('The response content must be 'ContentNotRenderedError: The response content must be rendered before it can be iterated over.

我发现这个set of notes描述了你在脚下射击以使用基于类的视图，为什么会这样？

编辑：我最终使用了这个...但我仍然希望有人能告诉我如何获得原始oneliner或类似的工作

class Handler500(TemplateView):
    template_name = "500.html"  
    @classmethod
    def as_error_view(cls):
        v = cls.as_view()
        def view(request):
            r = v(request)
            r.render()
            return r
        return view
handler500 = Handler500.as_error_view()

Answer 1

我宁愿在vanilla Django中使用带有静态HTML的库存500模板，然后对代码执行任何操作。这是一个我认为不应该触及的切换。

Answer 2

我认为它实际上非常简单（在Django 1.7和Python 3.4中）：

views.py

from django.http import HttpResponse
from django.views.generic.base import View

class Custom500View(View):
    def dispatch(self, request, *args, **kwargs):
        return HttpResponse('My custom django 500 page')

urls.py

from .views import Custom500View

handler500 = Custom500View.as_view()

Answer 3

你所需要的一切 -

def redirect_500_error(request):
    return render_to_response('errors/500.html', {}, context_instance=RequestContext(request))

Django handler500作为基于类的视图

3 个答案:

views.py

urls.py