Question

我在理解新CBV如何工作方面遇到了一些麻烦。我的问题是，我需要登录所有视图，其中一些是特定权限。在基于函数的视图中，我使用@permission_required（）和视图中的login_required属性执行此操作，但我不知道如何在新视图上执行此操作。 django文档中是否有一些部分解释了这一点？我没找到任何东西。我的代码出了什么问题？

我尝试使用@method_decorator，但它回复“ / errors / prueba / _wrapped_view（）中的TypeError至少需要1个参数（0给定）”

这是代码（GPL）：

from django.utils.decorators import method_decorator
from django.contrib.auth.decorators import login_required, permission_required

class ViewSpaceIndex(DetailView):

    """
    Show the index page of a space. Get various extra contexts to get the
    information for that space.

    The get_object method searches in the user 'spaces' field if the current
    space is allowed, if not, he is redirected to a 'nor allowed' page. 
    """
    context_object_name = 'get_place'
    template_name = 'spaces/space_index.html'

    @method_decorator(login_required)
    def get_object(self):
        space_name = self.kwargs['space_name']

        for i in self.request.user.profile.spaces.all():
            if i.url == space_name:
                return get_object_or_404(Space, url = space_name)

        self.template_name = 'not_allowed.html'
        return get_object_or_404(Space, url = space_name)

    # Get extra context data
    def get_context_data(self, **kwargs):
        context = super(ViewSpaceIndex, self).get_context_data(**kwargs)
        place = get_object_or_404(Space, url=self.kwargs['space_name'])
        context['entities'] = Entity.objects.filter(space=place.id)
        context['documents'] = Document.objects.filter(space=place.id)
        context['proposals'] = Proposal.objects.filter(space=place.id).order_by('-pub_date')
        context['publication'] = Post.objects.filter(post_space=place.id).order_by('-post_pubdate')
        return context

Answer 1

the CBV docs中列出了一些策略：

Add the decorator in your urls.py route，例如login_required(ViewSpaceIndex.as_view(..))

Decorate your CBV's dispatch method with a method_decorator例如，

from django.utils.decorators import method_decorator

@method_decorator(login_required, name='dispatch')
class ViewSpaceIndex(TemplateView):
    template_name = 'secret.html'

在Django 1.9之前，你不能在课堂上使用method_decorator，所以你必须覆盖dispatch方法：

class ViewSpaceIndex(TemplateView):

    @method_decorator(login_required)
    def dispatch(self, *args, **kwargs):
        return super(ViewSpaceIndex, self).dispatch(*args, **kwargs)

使用Django 1.9+中提供的django.contrib.auth.mixins.LoginRequiredMixin之类的访问混合，并在其他答案中概述：

from django.contrib.auth.mixins import LoginRequiredMixin

class MyView(LoginRequiredMixin, View):

    login_url = '/login/'
    redirect_field_name = 'redirect_to'

您收到TypeError的原因在文档中说明了：

注意： method_decorator将* args和** kwargs作为参数传递给类的修饰方法。如果您的方法不接受兼容的参数集，则会引发TypeError异常。

Answer 2

这是我的方法，我创建了一个受保护的mixin（这保存在我的mixin库中）：

from django.contrib.auth.decorators import login_required
from django.utils.decorators import method_decorator

class LoginRequiredMixin(object):
    @method_decorator(login_required)
    def dispatch(self, request, *args, **kwargs):
        return super(LoginRequiredMixin, self).dispatch(request, *args, **kwargs)

每当您想要保护视图时，只需添加适当的mixin：

class SomeProtectedViewView(LoginRequiredMixin, TemplateView):
    template_name = 'index.html'

请确保您的mixin是第一个。

更新：我在2011年发布了这个版本，从版本1.9开始，Django现在包括这个和其他有用的mixin（AccessMixin，PermissionRequiredMixin，UserPassesTestMixin）作为标准！

Answer 3

这是使用基于类的装饰器的替代方法：

from django.utils.decorators import method_decorator

def class_view_decorator(function_decorator):
    """Convert a function based decorator into a class based decorator usable
    on class based Views.

    Can't subclass the `View` as it breaks inheritance (super in particular),
    so we monkey-patch instead.
    """

    def simple_decorator(View):
        View.dispatch = method_decorator(function_decorator)(View.dispatch)
        return View

    return simple_decorator

然后可以像这样使用：

@class_view_decorator(login_required)
class MyView(View):
    # this view now decorated

Answer 4

我意识到这个帖子有点陈旧，但无论如何这里是我的两分钱。

使用以下代码：

from django.utils.decorators import method_decorator
from inspect import isfunction

class _cbv_decorate(object):
    def __init__(self, dec):
        self.dec = method_decorator(dec)

    def __call__(self, obj):
        obj.dispatch = self.dec(obj.dispatch)
        return obj

def patch_view_decorator(dec):
    def _conditional(view):
        if isfunction(view):
            return dec(view)

        return _cbv_decorate(dec)(view)

    return _conditional

我们现在有办法修补装饰器，因此它将变得多功能化。这实际上意味着当应用于常规视图装饰器时，如下所示：

login_required = patch_view_decorator(login_required)

这个装饰器在按照最初预期的方式使用时仍然可以工作：

@login_required
def foo(request):
    return HttpResponse('bar')

但在使用时也会正常工作：

@login_required
class FooView(DetailView):
    model = Foo

这似乎在我最近遇到的几种情况下都能正常工作，包括这个现实世界的例子：

@patch_view_decorator
def ajax_view(view):
    def _inner(request, *args, **kwargs):
        if request.is_ajax():
            return view(request, *args, **kwargs)
        else:
            raise Http404

    return _inner

编写ajax_view函数是为了修改（基于函数的）视图，因此只要非ajax调用访问此视图，就会引发404错误。通过简单地将补丁函数应用为装饰器，此装饰器也可以在基于类的视图中工作

Answer 5

对于那些使用 Django＆gt; = 1.9 的人，它已作为AccessMixin，LoginRequiredMixin，PermissionRequiredMixin包含在django.contrib.auth.mixins中UserPassesTestMixin

所以要将LoginRequired应用于CBV（例如DetailView）：

from django.contrib.auth.mixins import LoginRequiredMixin
from django.views.generic.detail import DetailView


class ViewSpaceIndex(LoginRequiredMixin, DetailView):
    model = Space
    template_name = 'spaces/space_index.html'
    login_url = '/login/'
    redirect_field_name = 'redirect_to'

记住GCBV Mixin顺序也很好： Mixins 必须在左侧侧进行，基本视图类必须去在正确方面。如果订单不同，您可能会遇到破碎和不可预测的结果。

Answer 6

如果它是大多数页面要求用户登录的站点，您可以使用中间件强制登录所有视图，除非特别标记的某些页面。

Pre Django 1.10 middleware.py:

from django.contrib.auth.decorators import login_required
from django.conf import settings

EXEMPT_URL_PREFIXES = getattr(settings, 'LOGIN_EXEMPT_URL_PREFIXES', ())

class LoginRequiredMiddleware(object):
    def process_view(self, request, view_func, view_args, view_kwargs):
        path = request.path
        for exempt_url_prefix in EXEMPT_URL_PREFIXES:
            if path.startswith(exempt_url_prefix):
                return None
        is_login_required = getattr(view_func, 'login_required', True)
        if not is_login_required:
            return None
        return login_required(view_func)(request, *view_args, **view_kwargs)

views.py：

def public(request, *args, **kwargs):
    ...
public.login_required = False

class PublicView(View):
    ...
public_view = PublicView.as_view()
public_view.login_required = False

您可以在设置中将您不想换行的第三方视图排除在外：

settings.py：

LOGIN_EXEMPT_URL_PREFIXES = ('/login/', '/reset_password/')

Answer 7

使用Django Braces。它提供了许多易于获得的有用混合物。它有漂亮的文档。试试看。

您甚至可以创建自定义混音。

http://django-braces.readthedocs.org/en/v1.4.0/

示例代码：

from django.views.generic import TemplateView

from braces.views import LoginRequiredMixin


class SomeSecretView(LoginRequiredMixin, TemplateView):
    template_name = "path/to/template.html"

    #optional
    login_url = "/signup/"
    redirect_field_name = "hollaback"
    raise_exception = True

    def get(self, request):
        return self.render_to_response({})

Answer 8

在我的代码中，我编写了这个适配器，以使成员函数适应非成员函数：

from functools import wraps


def method_decorator_adaptor(adapt_to, *decorator_args, **decorator_kwargs):
    def decorator_outer(func):
        @wraps(func)
        def decorator(self, *args, **kwargs):
            @adapt_to(*decorator_args, **decorator_kwargs)
            def adaptor(*args, **kwargs):
                return func(self, *args, **kwargs)
            return adaptor(*args, **kwargs)
        return decorator
    return decorator_outer

您可以像这样使用它：

from django.http import HttpResponse
from django.views.generic import View
from django.contrib.auth.decorators import permission_required
from some.where import method_decorator_adaptor


class MyView(View):
    @method_decorator_adaptor(permission_required, 'someapp.somepermission')
    def get(self, request):
        # <view logic>
        return HttpResponse('result')

Answer 9

使用django＆gt;这非常简单1.9支持PermissionRequiredMixin和LoginRequiredMixin

只需从身份验证

导入即可

views.py

from django.contrib.auth.mixins import LoginRequiredMixin

class YourListView(LoginRequiredMixin, Views):
    pass

有关详细信息，请参阅Authorization in django

Answer 10

如果您正在执行需要各种权限测试的项目，则可以继承此类。

from django.contrib.auth.decorators import login_required
from django.contrib.auth.decorators import user_passes_test
from django.views.generic import View
from django.utils.decorators import method_decorator



class UserPassesTest(View):

    '''
    Abstract base class for all views which require permission check.
    '''


    requires_login = True
    requires_superuser = False
    login_url = '/login/'

    permission_checker = None
    # Pass your custom decorator to the 'permission_checker'
    # If you have a custom permission test


    @method_decorator(self.get_permission())
    def dispatch(self, *args, **kwargs):
        return super(UserPassesTest, self).dispatch(*args, **kwargs)


    def get_permission(self):

        '''
        Returns the decorator for permission check
        '''

        if self.permission_checker:
            return self.permission_checker

        if requires_superuser and not self.requires_login:
            raise RuntimeError((
                'You have assigned requires_login as False'
                'and requires_superuser as True.'
                "  Don't do that!"
            ))

        elif requires_login and not requires_superuser:
            return login_required(login_url=self.login_url)

        elif requires_superuser:
            return user_passes_test(lambda u:u.is_superuser,
                                    login_url=self.login_url)

        else:
            return user_passes_test(lambda u:True)

Answer 11

我已根据Josh的解决方案进行了修复

class LoginRequiredMixin(object):

    @method_decorator(login_required)
    def dispatch(self, *args, **kwargs):
        return super(LoginRequiredMixin, self).dispatch(*args, **kwargs)

样本用法：

class EventsListView(LoginRequiredMixin, ListView):

    template_name = "events/list_events.html"
    model = Event

Answer 12

已经有一段时间了，现在Django发生了很大变化。

在此处检查如何装饰基于类的视图。

https://docs.djangoproject.com/en/2.2/topics/class-based-views/intro/#decorating-the-class

文档中没有包含“带有任何参数的装饰器”的示例。但是带有参数的装饰器是这样的：

def mydec(arg1):
    def decorator(func):
         def decorated(*args, **kwargs):
             return func(*args, **kwargs) + arg1
         return decorated
    return deocrator

因此，如果我们要将mydec用作不带参数的“常规”装饰器，则可以这样做：

mydecorator = mydec(10)

@mydecorator
def myfunc():
    return 5

类似地，将permission_required与method_decorator一起使用

我们可以做到：

@method_decorator(permission_required("polls.can_vote"), name="dispatch")
class MyView:
    def get(self, request):
        # ...

Answer 13

这是Permission_required装饰器的解决方案：

class CustomerDetailView(generics.GenericAPIView):

@method_decorator(permission_required('app_name.permission_codename', raise_exception=True))
    def post(self, request):
        # code...
        return True

如何在基于django类的视图上使用permission_required装饰器

13 个答案: