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Home/ Questions/Q 4026738
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T11:00:33+00:00 2026-05-20T11:00:33+00:00

I an writing a website that will have multiple skins. Each skin has its

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I an writing a website that will have multiple skins. Each skin has its own style sheet. I would like the skin used to be determined by a GET variable, so that this URL:

whatever?skin=foo

will cause the page to be rendered containing this HTML in the header:

<link rel="stylesheet" type="text/css" href="/site_media/foo.css"/>

(Normally I want the skin to be determined by a user’s preference, but I want this way of doing it as well, so a user can preview what a new skin will look like, as well as for making it easy for me while developing it.)

This is fairly easy to do in Django, for example you could use a template with this line:

<link rel="stylesheet" type="text/css" href="/site_media/{{skin}}.css"/>

And a view like this:

def whateverView(request):
    """ called by URL /whatever """
    skin = request.GET.get('skin', "default")
    c = RequestContext(request, {'skin': skin})
    html = whateverTemplate.render(c)
    return HttpResponse(html)

But I don’t want to have to do it like that, as I would have to add the same code to every single view, which would violate DRY.

So is there some way I can do it, such that it works on all my pages, while only writing the code once?

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  1. Editorial Team
    Editorial Team
    2026-05-20T11:00:34+00:00Added an answer on May 20, 2026 at 11:00 am

    You can do this using Django’s Context Processors: http://docs.djangoproject.com/en/dev/ref/templates/api/?#writing-your-own-context-processors. Or, you could enable django.core.conext_processors.request and access the request object in your template.

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