从Django模板中的URL访问kwarg [英] Access kwargs from a URL in a Django template

问题描述

我可以在Django模板中访问命名参数的值（从URL）吗？

Can I access value of a named argument (from the URL) in a Django template?

就像我可以访问 this_name的值一样从下面的Django模板中获取？

Like can I access the value of this_name below from a django template?

url(r'^area/(?P<this_name>[\w-]+)/$', views.AreaView.as_view(), name="area_list")

我可以获取整个URL路径并将其分解，但是想检查是否有一种直接的方法，因为它已经有了一个名称。

I could get the whole URL path and break it up but wanted to check if there's a straight forward way to do that, since it already has a name.

将其从视图中传递到上下文数据中可能是另一种选择，但是不确定我是否确实需要将其传递出去，因为我猜想模板已经具有不知何故？在请求API <中找不到直接方法/ a>。

Passing it down in the context data from the view may be an alternative but not sure if I do need to pass it down since I'd guess the template would already have it somehow? Couldn't find a direct method in the request API though.

推荐答案

在视图中，您可以访问URL args 和 kwargs 作为 self.args 和 self.kwargs 。

In the view, you can access the URL args and kwargs as self.args and self.kwargs.

class MyView(View):
    def my_method(self):
        this_name = self.kwargs['this_name']

如果您只想访问模板中的值，则您无需在视图中进行任何更改。基本 get_context_data 方法将视图作为 view 添加到上下文中，因此您可以将以下内容添加到模板中：

If you only want to access the value in the template, then you don't need to make any changes in the view. The base get_context_data method adds the view to the context as view, so you can add the following to your template:

{{ view.kwargs.this_name }}

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