Question

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

@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User和一个String。我想传递请求对象是否可能？另外，任何人都可以告诉我，我们如何直接访问装饰器内的User对象。

Answer 1

不，您无法将请求传递给user_passes_test。要了解其原因和方法，请转到source：

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

这是user_passes_test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，lambda u: has_add_permission(u, "project")）仅传递一个参数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只修改它）也可以传递request本身，但是您无法使用默认{ {1}}实施。

Answer 2

请注意，Django 1.9引入了UserPassesTestMixin，它使用方法test_func作为测试函数。这意味着请求在self.request中可用。所以你可以这样做：

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

但这仅适用于基于类的视图。

Answer 3

我发现编辑user_passes_test让装饰函数在request上运行而非request.user不会过于困难。我在this blog post中有一个关于视图装饰器装饰器的简短版本，但是对于后代，这里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

我唯一做的就是将if test_func(request.user):更改为if test_func(request):。

如何在user_passes_test装饰器可调用函数中传递Django请求对象

3 个答案: