ViewSet方法的Django rest框架permission_classes

Question

我正在使用 Django REST 框架编写一个 REST API，我想通过权限保护某些端点。权限类看起来像是提供了一种优雅的方式来实现这一点。我的问题是我想为不同的重写 ViewSet 方法使用不同的权限类。

class UserViewSet(viewsets.ModelViewSet):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    def create(self, request, *args, **kwargs):
        return super(UserViewSet, self).create(request, *args, **kwargs)

    @decorators.permission_classes(permissions.IsAdminUser)
    def list(self, request, *args, **kwargs):
        return super(UserViewSet, self).list(request, *args, **kwargs)

在上面的代码中，我也想允许未经身份验证的用户注册（创建用户），但我不想让任何人都可以列出用户，只为员工。

在docs 中，我看到了使用permission_classes 装饰器保护API 视图（不是ViewSet 方法）的示例，并且我看到了为整个ViewSet 设置权限类。但它似乎不适用于覆盖的 ViewSet 方法。有没有办法只将它们用于某些端点？

Answer 1

我认为没有内置的解决方案。但是您可以通过覆盖get_permissions 方法来实现这一点：

from rest_framework.permissions import AllowAny, IsAdminUser

class UserViewSet(viewsets.ModelViewSet):
    queryset = User.objects.all()
    serializer_class = UserSerializer

    permission_classes_by_action = {'create': [AllowAny],
                                    'list': [IsAdminUser]}

    def create(self, request, *args, **kwargs):
        return super(UserViewSet, self).create(request, *args, **kwargs)

    def list(self, request, *args, **kwargs):
        return super(UserViewSet, self).list(request, *args, **kwargs)

    def get_permissions(self):
        try:
            # return permission_classes depending on `action` 
            return [permission() for permission in self.permission_classes_by_action[self.action]]
        except KeyError: 
            # action is not set return default permission_classes
            return [permission() for permission in self.permission_classes]

Answer 2

我创建了一个派生自@ilse2005 答案的超类。在所有后续的 django 视图中你可以继承这个来实现动作级别的权限控制。

class MixedPermissionModelViewSet(viewsets.ModelViewSet):
   '''
   Mixed permission base model allowing for action level
   permission control. Subclasses may define their permissions
   by creating a 'permission_classes_by_action' variable.

   Example:
   permission_classes_by_action = {'list': [AllowAny],
                                   'create': [IsAdminUser]}
   '''

   permission_classes_by_action = {}

   def get_permissions(self):
      try:
        # return permission_classes depending on `action`
        return [permission() for permission in self.permission_classes_by_action[self.action]]
      except KeyError:
        # action is not set return default permission_classes
        return [permission() for permission in self.permission_classes]

Answer 3

我可能迟到了，但正如其中一位评论者指出的那样，我使用了 mixin。从@Itachi 那里得到答案，这是我的 mixin 实现：

class ViewSetActionPermissionMixin:
    def get_permissions(self):
        """Return the permission classes based on action.

        Look for permission classes in a dict mapping action to
        permission classes array, ie.:

        class MyViewSet(ViewSetActionPermissionMixin, ViewSet):
            ...
            permission_classes = [AllowAny]
            permission_action_classes = {
                'list': [IsAuthenticated]
                'create': [IsAdminUser]
                'my_action': [MyCustomPermission]
            }

            @action(...)
            def my_action:
                ...

        If there is no action in the dict mapping, then the default
        permission_classes is returned. If a custom action has its
        permission_classes defined in the action decorator, then that
        supercedes the value defined in the dict mapping.
        """
        try:
            return [
                permission()
                for permission in self.permission_action_classes[self.action]
            ]
        except KeyError:
            if self.action:
                action_func = getattr(self, self.action, {})
                action_func_kwargs = getattr(action_func, "kwargs", {})
                permission_classes = action_func_kwargs.get(
                    "permission_classes"
                )
            else:
                permission_classes = None

            return [
                permission()
                for permission in (
                    permission_classes or self.permission_classes
                )
            ]

下面是 mixin 的使用方法：

class MyViewSet(ViewSetActionPermissionMixin, ModelViewSet):
    ...
    permission_action_classes = {
        "list": [AllowAny],
        "create": [IsAdminUser],
        "custom_action": [MyCustomPermission],
    }

    @action(...)
    def custom_action(self, request, *args, **kwargs):
        ...

Answer 4

我认为所有其他答案都很棒，但我们不应该直接取消在装饰器中定义的默认操作 permission_classes。所以，

from rest_framework import viewsets
from rest_framework import permissions

class BaseModelViewSet(viewsets.ModelViewSet):
    queryset = ''
    serializer_class = ''
    permission_classes = (permissions.AllowAny,)

    # Refer to https://stackoverflow.com/a/35987077/1677041
    permission_classes_by_action = {
        'create': permission_classes,
        'list': permission_classes,
        'retrieve': permission_classes,
        'update': permission_classes,
        'destroy': permission_classes,
    }

    def get_permissions(self):
        try:
            return [permission() for permission in self.permission_classes_by_action[self.action]]
        except KeyError:
            if self.action:
                action_func = getattr(self, self.action, {})
                action_func_kwargs = getattr(action_func, 'kwargs', {})
                permission_classes = action_func_kwargs.get('permission_classes')
            else:
                permission_classes = None

            return [permission() for permission in (permission_classes or self.permission_classes)]

现在我们可以用这两种方式定义permission_classes。由于我们在超类中定义了默认的全局 permission_classes_by_action，因此我们可以为选项 2 中的所有操作删除该定义。

class EntityViewSet(BaseModelViewSet):
    """EntityViewSet"""
    queryset = Entity.objects.all()
    serializer_class = EntitySerializer
    permission_classes_by_action = {
        'create': (permissions.IsAdminUser,),
        'list': (permissions.IsAuthenticatedOrReadOnly,),
        'retrieve': (permissions.AllowAny,),
        'update': (permissions.AllowAny,),
        'destroy': (permissions.IsAdminUser,),
        'search': (permissions.IsAuthenticated,)  # <--- Option 1
    }

    @action(detail=False, methods=['post'], permission_classes=(permissions.IsAuthenticated,))  # <--- Option 2
    def search(self, request, format=None):
        pass