21

我正在使用 Django user_passes_test 装饰器来检查用户权限。

@user_passes_test(lambda u: has_add_permission(u, "project"))
def create_project(request):
......

我正在调用一个回调函数 has_add_permission，它接受两个参数用户和一个字符串。我想将请求对象连同它一起传递，这可能吗？另外，谁能告诉我我们如何能够直接访问装饰器内的用户对象。

4

3 回答 3

15

不，您不能将请求传递给user_passes_test. 要了解它为什么以及如何工作，只需转到源代码：

def user_passes_test(test_func, login_url=None, redirect_field_name=REDIRECT_FIELD_NAME):
    """
    Decorator for views that checks that the user passes the given test,
    redirecting to the log-in page if necessary. The test should be a callable
    that takes the user object and returns True if the user passes.
    """

    def decorator(view_func):
        @wraps(view_func, assigned=available_attrs(view_func))
        def _wrapped_view(request, *args, **kwargs):
            if test_func(request.user):
                return view_func(request, *args, **kwargs)
            path = request.build_absolute_uri()
            # If the login url is the same scheme and net location then just
            # use the path as the "next" url.
            login_scheme, login_netloc = urlparse.urlparse(login_url or
                                                        settings.LOGIN_URL)[:2]
            current_scheme, current_netloc = urlparse.urlparse(path)[:2]
            if ((not login_scheme or login_scheme == current_scheme) and
                (not login_netloc or login_netloc == current_netloc)):
                path = request.get_full_path()
            from django.contrib.auth.views import redirect_to_login
            return redirect_to_login(path, login_url, redirect_field_name)
        return _wrapped_view
    return decorator

user_passes_test这是装饰器背后的代码。如您所见，传递给装饰器的测试函数（在您的情况下为lambda u: has_add_permission(u, "project")）仅传递了一个参数request.user. 现在，当然可以编写自己的装饰器（甚至直接复制此代码并对其进行修改）也可以传递request自身，但您不能使用默认user_passes_test实现来做到这一点。

于 2012-08-08T20:48:37.817 回答
11

请注意，Django 1.9 引入了UserPassesTestMixin，它使用方法test_func作为测试函数。这意味着该请求在self.request. 所以你可以做这样的事情：

class MyView(UserPassesTestMixin, View):
    def test_func(self):
        return has_add_permission(self.request.user, self.request)

然而，这只适用于基于类的视图。

于 2016-02-27T20:14:01.680 回答
5

我发现编辑user_passes_test让装饰功能运行request而不是request.user过于困难。我在这篇博客文章中有一个关于视图装饰器装饰器的简短版本，但为了后代，这是我完整编辑的代码：

def request_passes_test(test_func, login_url=None, redirect_field_name=REDIRECT_FIELD_NAME):
    """
    Decorator for views that checks that the request passes the given test,
    redirecting to the log-in page if necessary. The test should be a callable
    that takes the request object and returns True if the request passes.
    """

    def decorator(view_func):
        @wraps(view_func, assigned=available_attrs(view_func))
        def _wrapped_view(request, *args, **kwargs):
            if test_func(request):
                return view_func(request, *args, **kwargs)
            path = request.build_absolute_uri()
            # urlparse chokes on lazy objects in Python 3, force to str
            resolved_login_url = force_str(
                resolve_url(login_url or settings.LOGIN_URL))
            # If the login url is the same scheme and net location then just
            # use the path as the "next" url.
            login_scheme, login_netloc = urlparse(resolved_login_url)[:2]
            current_scheme, current_netloc = urlparse(path)[:2]
            if ((not login_scheme or login_scheme == current_scheme) and
                    (not login_netloc or login_netloc == current_netloc)):
                path = request.get_full_path()
            from django.contrib.auth.views import redirect_to_login
            return redirect_to_login(
                path, resolved_login_url, redirect_field_name)
        return _wrapped_view
    return decorator

几乎我唯一做的就是更改if test_func(request.user):为if test_func(request):.

于 2014-06-26T06:35:21.650 回答