Hello!
You need create a "static" folder in your project and inside in this folder
you create a "css" folder.
I use this configuration and works.
settings.py
import os.path
STATIC_ROOT = ''
# URL prefix for static files.
STATIC_URL = '/static/'
ADMIN_MEDIA_PREFIX = '/static/admin/'
TEMPLATE_DIRS = (
os.path.join(os.path.dirname(__file__), 'templates'),
)
And call a css for example, in base.html in this way:
<link href="{{ STATIC_URL }}css/style.css" rel="stylesheet" type="text/css"
media="screen" />
Regards.
2011/4/25 Joakim Hove <joakim.h...@gmail.com>
> Hello,
>
> I have just started using the {{ STATIC_URL }} template tag to insert
> proper url's to static resources in my templates. To use this tag I
> need to create the context for rendering as RequestContext() instance,
> which takes the request as a parameter for the initialisation.
>
> In my code I have several methods attached to models which create a
> suitable context for rendering that model instance in a specfied way:
>
> class MyModel( models.Model ):
> ....
> ....
>
> def create_xxx_context( self , **kwargs):
> ....
>
> def create_yyy_context( self, ,**kwargs):
> ....
>
> (Maybe that is a horrific design in the first place ???).
>
>
> Anyway - the methods create_???_context() do not have natural access
> to the request object (and I am reluctant to pass that argument on) -
> so I was wondering if I could achieve exactly the same by just
> "manually" adding the context variable STATIC_URL like:
>
> from django.conf import settings
> ....
>
> def view(request, args):
> context = .....
> context["STATIC_URL"] = settings.STATIC_URL
> return render_to_response( template , context )
>
> Or is there more magic to the {{ STATIC_URL }} tag when it comes as a
> RequestContext()?
>
>
> Joakim
>
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