[英]Pyramid and FormAlchemy admin interface
我有一个使用formalchemy管理界面的金字塔项目。 我添加了基本的ACL身份验证,即使我通过了身份验证,pyramid_formalchemy插件总是会拒绝。
有关如何仅允许经过身份验证的用户使用pyramid_formalchemy管理界面的任何想法?
授权策略添加如下:
authn_policy = AuthTktAuthenticationPolicy('MYhiddenSECRET', callback=groupfinder) authz_policy = ACLAuthorizationPolicy() config = Configurator( settings=settings, root_factory='package.auth.RootFactory', authentication_policy=authn_policy, authorization_policy=authz_policy ) # pyramid_formalchemy's configuration config.include('pyramid_formalchemy') config.include('fa.jquery') config.formalchemy_admin('admin', package='package', view='fa.jquery.pyramid.ModelView')
pyramid_formalchemy
使用权限'view', 'edit', 'delete', 'new'
来确定谁可以做什么。 __acl__
从SQLAlchemy模型对象向下传播。 因此,您需要在每个模型对象上放置__acl__
,以允许所需的组访问这些权限。 例如,来自pyramid_formalchemy
pyramidapp
示例项目:
class Bar(Base):
__tablename__ = 'bar'
__acl__ = [
(Allow, 'admin', ALL_PERMISSIONS),
(Allow, 'bar_manager', ('view', 'new', 'edit', 'delete')),
]
id = Column(Integer, primary_key=True)
foo = Column(Unicode(255))
当然,如果你不提供__acl__
那么它将会查看资源树的谱系,直到它到达factory
。 默认情况下, pyramid_formalchemy
定义了自己的工厂pyramid_formalchemy.resources.Models
,但是您可以将其子类化并为其提供__acl__
,作为所有模型的全局:
from pyramid_formalchemy.resources import Models
class ModelsWithACL(Models):
"""A factory to override the default security setting"""
__acl__ = [
(Allow, 'admin', ALL_PERMISSIONS),
(Allow, Authenticated, 'view'),
(Allow, 'editor', 'edit'),
(Allow, 'manager', ('new', 'edit', 'delete')),
]
config.formalchemy_admin('admin', package='package', view=..., factory=ModelsWithACL)
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