Question

我尝试在迁移中创建自定义权限，但是在运行迁移后，未在权限表中创建权限。有人能指出错误是什么吗？此外，我不确定我应该使用什么作为ContentType的相关模型，因为该权限用于限制可以查看显示网站上用户摘要的页面的用户。非常感谢任何帮助，谢谢。

def add_view_aggregated_data_permissions(apps, schema_editor):
    ContentType = apps.get_model('django', 'ContentType')
    Permission = apps.get_model('auth', 'Permission')
    content_type = ContentType.objects.get(app_label='auth', model='user')
    permission = Permission.objects.create(codename='can_view_data',
                                           name='Can view data',
                                           content_type=content_type)

Answer 1

我建议您使用标准方式来使用Django documentation中所述的自定义权限。你将完全避免许多问题。

要为给定模型对象创建自定义权限，请使用权限模型Meta属性。

此示例模型创建自定义权限：

class MyModel(models.Model):
    ...
    class Meta:
        permissions = (
            ('view_data', "Can see available data"),
        )

这样做的唯一方法就是在运行manage.py migrate时创建额外的权限。当用户试图访问应用程序提供的功能时，您的代码负责检查这些权限的值...

然后，您可以在视图中使用permission_required装饰器来检查特定权限：

from django.contrib.auth.decorators import permission_required

@permission_required('myapp.view_data')
def my_view(request):
    ...

Answer 2

从django 1.8和内置迁移开始，这是非常无痛的。

您需要做的就是将相关权限添加到相关的权限中模型
运行makemigration

./ manage.py makemigrations
运行上一步中创建的迁移

./ manage.py migrate

Answer 3

我想为所有应用模型创建自定义权限（读取）。我做了两个步骤：

从DjangoModelPermissions创建扩展权限：

class DjangoModelPermissionsExtended(DjangoModelPermissions):
    """
    """
    perms_map = {
        'GET': ['%(app_label)s.read_%(model_name)s'],
        'OPTIONS': [],
        'HEAD': [],
        'POST': ['%(app_label)s.add_%(model_name)s'],
        'PUT': ['%(app_label)s.change_%(model_name)s'],
        'PATCH': ['%(app_label)s.change_%(model_name)s'],
        'DELETE': ['%(app_label)s.delete_%(model_name)s'],
    }

将其置于每个视图中我想拥有读取权限：

class ExampleViewSet(viewsets.ModelViewSet):
    permission_classes = (
        DjangoModelPermissionsExtended,
    )

创建django command customread.py：

from django.core.management.base import BaseCommand, CommandError
from project.app import models as app_models
from django.db import models
from django.contrib.auth.models import Permission
from django.contrib.contenttypes.models import ContentType
import inspect

class Command(BaseCommand):
help = 'Create the read permission to app models'

def handle(self, *args, **options):
    for name, obj in inspect.getmembers(app_models):
        if inspect.isclass(obj) and issubclass(obj, models.Model):
            try:
                self.add_canread(obj)
                self.stdout.write(self.style.SUCCESS(
                    'created permission for %s' % obj
                ))
            except Exception as e:
                self.stdout.write(self.style.ERROR(
                    'Permission already exists for %s' % obj
                ))

def add_canread(self, object_class):
    """This a function that can be executed in order to create
    new permissions (read view) to a class in DB.

    """
    if inspect.isclass(object_class):
        content_type = ContentType.objects.get_for_model(object_class)
        permission = Permission.objects.create(
            codename='read_{}'.format(object_class._meta.model_name),
            name='Can view {}'.format(object_class.__name__),
            content_type=content_type,
        )
    else:
        msg = "The object is not a class"
        print(msg)

执行迁移后执行：
```
python manage.py customread
```

在数据迁移中创建自定义权限

3 个答案: