如何从Django中的urls.py访问HttpRequest [英] How to access HttpRequest from urls.py in Django

问题描述

基本上我想使用一个通用的视图来根据用户名列出对象。现在的问题是，我该怎么做，如下所示：

Basically I want to use a generic view that lists objects based on a username. Now, the question is, how do I do something like:

(r'^resources/$',
  ListView.as_view(
    queryset=Resources.objects.filter(user=request.user.username),
    ...
  )
)

我找不到一种方法来访问HttpRequest（请求）对象，但是...或者我需要使用我自己的视图做所有的对象选择？

I couldn't find a way to access the HttpRequest (request) object though... Or do I need to use my own views and do all object selection there?

推荐答案

如果你真的想直接混淆你的URLconf，你可以这样做： p>

If you really want to clutter your URLconf directly, you can do it like so:

(r'^resources/$',
 lambda request: ListView.as_view(queryset=Resources.objects.filter(user=request.user.username), ...)(request)
)

或者通过子视图来访问请求：

Or access the request by subclassing the view:

class MyListView(ListView):
    def dispatch(self, request, *args, **kwargs):
        self.queryset = Resources.objects.filter(user = request.user.username)
        return super(MyListView, self).dispatch(request, *args, **kwargs)

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