41

我有以下 3 个课程:

class Resource:
    id = Column(Integer, primary_key=True)
    path = Column(Text)
    data = Column(Binary)
    type = Column(Text)

    def set_resource(self, path, data, type):
        self.path = path
        self.data = data
        self.type = type

class EnvironmentResource(Base, Resource):
    __tablename__ = 'environment_resources'
    parent_id = Column(Integer, ForeignKey('environments.id', ondelete='CASCADE'))
    def __init__(self, path, data, type):
        self.set_resource(path, data, type)

class Environment(Base):
    __tablename__ = 'environments'
    id = Column(Integer, primary_key=True)
    identifier = Column(Text, unique=True)
    name = Column(Text)
    description = Column(Text)

    _resources = relationship("EnvironmentResource",
        cascade="all, delete-orphan",
        passive_deletes=True)
    _tools = relationship("Tool",
        cascade="all, delete-orphan",
        passive_deletes=True)

    def __init__(self, name, identifier, description):
        self.name = name
        self.identifier = identifier
        self.description = description

    def get_resource(self, path):
        return self._resources.filter(EnvironmentResource.path==path).first()

在调用 get_resource 时,我被告知 'InstrumentedList' 对象没有属性 'filter' - 我已经阅读了文档并且无法完全弄清楚这一点。我错过了什么,以便我可以过滤与我的“get_resource”方法中的环境相对应的资源?

PS:我知道 get_resource 会抛出异常,这就是我希望它做的事情。

4

1 回答 1

77

为了像 with 一样使用关系Query,您需要将其配置为lazy='dynamic'。在动态关系加载器中查看更多信息:

_resources = relationship("EnvironmentResource",
    cascade="all, delete-orphan",
    lazy='dynamic',
    passive_deletes=True)
于 2012-07-20T12:24:43.533 回答