views:

1534

answers:

2

I'm trying to make a Preview function. I'm reading this blog, Django Admin Preview, but now I have the following error and I don't know what it means.

    Traceback (most recent call last):

     File "/home/user/webapps/django/lib/python2.5/django/core/handlers/base.py", line 92, in get_response
       response = callback(request, *callback_args, **callback_kwargs)

    TypeError: 'str' object is not callable

I'm lost..

Edit:

Thanks guys/gals, here is my view.py and url.py:

    from diligencia.diligencias.views import preview

    url(r'^admin/diligencias/diligencia/(?P<object_id>\d+)/preview/$','preview'),
    (r'^admin/(.*)', admin.site.root),

    from diligencia.diligencias.models import Diligencia

    @staff_member_required
    def preview(request, object_id):
        return object_detail(request, object_id=object_id,queryset=Diligencia.objects.all(), template_object_name = 'diligencia_detail.html', )
+1  A: 

I suspect your view isn't a function. Make sure the argument in your urls.py is a function that takes one parameter. Like :

import default

url(r'^s(?:ite)?/search$', default.search, name="search"),

And then you have in default.py

def search(request) :
    # do stuff
Paul Tarjan
+2  A: 

The signature for the url function within a urlconf is like this:

def url(regex, view, kwargs=None, name=None, prefix='')

You are using positional parameters only, but are passing only regex, view and name. So Python thinks your third parameter is the kwargs dictionary, not the name.

Instead, do this:

url(r'^admin/diligencias/diligencia/(?P<object_id>\d+)/preview/$', name='preview'),

to pass the name as a kwarg so that Python recognises it properly.

Daniel Roseman
Daniel Thanks, with a litter change is working: url(r'^admin/diligencias/diligencia/(?P<object_id>\d+)/preview/$', preview, name='preview'),
Asinox