django - 如何激活 DJANGO.CORE.CONTEXT_PROCESSORS.REQUEST

Question

我读了这个

DJANGO.CORE.CONTEXT_PROCESSORS.REQUEST

If TEMPLATE_CONTEXT_PROCESSORS contains this processor, every RequestContext will contain a variable request, which is the current HttpRequest. Note that this processor is not enabled by default; you'll have to activate it.

来自 this page

但似乎没有信息如何激活这个处理器。

这是我原来的问题

Access request in django custom template tags

在我按照答案之后

我仍然有错误

TemplateSyntaxError at / Caught an exception while rendering: 'request' Original Traceback (most recent call last): 
File "C:\Python25\lib\site-packages\django\template\debug.py", line 71, in render_node result = node.render(context) 
File "C:\Python25\lib\site-packages\django\template__init__.py", line 936, in render dict = func(*args)
 File "c:\...\myapp_extras.py", line 7, in login request = context['request'] 
File "C:\Python25\lib\site-packages\django\template\context.py", line 44, in getitem raise KeyError(key) KeyError: 'request'

导致问题的代码是

request = context['request'] 在

from django import template

register = template.Library()


@register.inclusion_tag('userinfo.html',takes_context = True)
def userinfo(context):
 request = context['request']
 address = request.session['address']
 return {'address':address}

Answer 1

我在这里回答:How can I pass data to any template from any view in Django?

另请参阅对我的回答的评论……您可能也需要这些信息。