如何在 Django 中使用多个样式表，其中样式表由 GET 变量确定，而不违反 DRY？
我正在编写一个具有多种皮肤的网站。每个皮肤都有自己的样式表。我希望皮肤由 GET 变量确定，这样这个 URL:
whatever?skin=foo
将导致页面被渲染，在标题中包含这个 HTML:
<link rel="stylesheet" type="text/css" href="/site_media/foo.css"/>
(通常我希望皮肤由用户的偏好确定，但我想要这种方式也是如此，这样用户就可以预览新皮肤的外观，并让我在开发它时变得更容易。）
这在 Django 中相当容易做到，例如您可以使用带有此行的模板：
<link rel="stylesheet" type="text/css" href="/site_media/{{skin}}.css"/>
和这样的视图：
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)
但我不想这样做，因为我必须向每个视图添加相同的代码，这会违反 干。
那么有什么方法可以做到这一点，使其适用于我的所有页面，同时只编写一次代码？
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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您可以使用 Django 的上下文处理器来完成此操作： http://docs.djangoproject.com/en/dev/ref/templates/api/?#writing-your-own-context-processors。或者，您可以启用 django.core.conext_processors.request 并访问模板中的请求对象。
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.requestand access the request object in your template.