在views.py中的urls.py中包装/装饰函数

Question

所以,我非常熟悉在views.py中包装函数.所以我写了一个装饰,如果用户重定向到默认REDIRECT_URL 是在(某种反向的记录login_required); 这是基于我过去的观点:

def not_logged_in(redirect_url=None, redirect_field_name=REDIRECT_FIELD_NAME):
    def decorator(view_func, *args, **kwargs):
        def wrapper(request, *args, **kwargs):
            if not request.user.is_authenticated():
                return view_func(*args, **kwargs)
            else:
                redirect_url = (request.REQUEST.get(redirect_field_name, redirect_url) or
                                settings.REDIRECT_URL)
                return HttpResponseRedirect(redirect_url)
        return wrapper
    return decorator

但是,我收到以下错误:'function' object has no attribute 'status_code'这是由MiddleWare期望HttpResponse引起的.当我response看到它的价值时,我看到了它<function wrapper at 0x2b3a9922a500>.

以下是我在调用它的方式urls.py:

url(r'login/', 
     not_logged_in(auth_views.login), 
     {'authentication_form': LoginForm },
),

Answer 1

这是我对同一件事的实现.

def logout_required(view):
    def f(request, *args, **kwargs):
        if request.user.is_anonymous():
            return view(request, *args, **kwargs)
        return HttpResponseRedirect(settings.LOGIN_REDIRECT_URL)
    return f

在urls.py:

urlpatterns = patterns("",
    url(r"^login/", logout_required(login), {"template_name": "users/login.html"}, "login"),
    # ...

我希望这有帮助(不确定).