On Tue, Nov 2, 2010 at 1:00 PM, sami nathan <itssami.g...@gmail.com> wrote:
> THIS IS HOW MY URL.PY LOOKS LIKE
>
> from django.conf.urls.defaults import *
> from it.view import current_datetime
>
>
>
> # Uncomment the next two lines to enable the admin:
> #from django.contrib import admin
> #admin.autodiscover()
>
>
> urlpatterns = patterns('',
> # Example:
> (r"^wap/di/sub/$",current_datetime),
> # Uncomment the admin/doc line below to enable admin documentation:
> # (r'^admin/doc/', include('django.contrib.admindocs.urls')),
>
> # Uncomment the next line to enable the admin:
> # (r'^admin/', include(admin.site.urls)),
> )
>
> `~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
> My view .py looks like this
>
> from django.http import *
> import urllib
> from django_restapi.responder import XMLResponder
>
>
> def current_datetime(request):
> word = request.GET['word']
> message =
> urllib.urlopen('http://m.broov.com/wap/di/sub?word='+word+'&type=00submit=Submit',)
> return XMLResponder(message)
>
> But my error is
> Exception Value:
>
> 'XMLResponder' object has no attribute 'status_code'
>
> Exception Location:
> D:\Python25\lib\site-packages\django\middleware\common.py in
> process_response, line 94
>
Judging from http://code.google.com/p/django-rest-interface/ you
aren't using the XMLResponder class at all correctly. As far as I can
tell, one is supposed to use XMLResponder to define a 'resource',
which is then provide as the callable in urls.py.
You are using it as a replacement for HttpResponse, which it isn't,
and because it isn't you get the error you describe.
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