Python-сообщество

Добрый день,

потихоньку читаю djangobook.

пытаюсь изобразить код из 9й главый.

from django.template import loader, RequestContext

def custom_proc(request):
    "A context processor that provides 'app', 'user' and 'ip_address'."
    return {
        'app': 'My app',
        'user': request.user,
        'ip_address': request.META['REMOTE_ADDR']
    }

def view_1(request):
    # ...
    t = loader.get_template('template1.html')
    c = RequestContext(request, {'message': 'I am view 1.'},
            processors=[custom_proc])
    return t.render(c)

Traceback (most recent call last):

File “/usr/lib/pymodules/python2.6/django/core/servers/basehttp.py”, line 279, in run
self.result = application(self.environ, self.start_response)

File “/usr/lib/pymodules/python2.6/django/core/servers/basehttp.py”, line 651, in __call__
return self.application(environ, start_response)

File “/usr/lib/pymodules/python2.6/django/core/handlers/wsgi.py”, line 245, in __call__
response = middleware_method(request, response)

File “/usr/lib/pymodules/python2.6/django/middleware/locale.py”, line 21, in process_response
patch_vary_headers(response, ('Accept-Language',))

File “/usr/lib/pymodules/python2.6/django/utils/cache.py”, line 130, in patch_vary_headers
if response.has_header('Vary'):

AttributeError: ‘SafeUnicode’ object has no attribute ‘has_header’

Отредактировано (Фев. 25, 2010 17:54:01)

В джангобуке ошибка - каждый вью должен возвращать HttpResponse, а не строку

return HttpResponse(t.render(c))

Благодарю покорно