Python sqlalchemy:“InstrumentedList”对象没有属性“filter”
我有以下三门课:Python sqlalchemy:“InstrumentedList”对象没有属性“filter”,python,sqlalchemy,Python,Sqlalchemy,我有以下三门课: 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 sel
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,不能完全理解这一点。我遗漏了什么,以便能够在“get_resource”方法中过滤与环境相对应的资源
PS:我知道get_资源将抛出一个异常,这是我希望它做的。为了使用as with查询,您需要使用lazy='dynamic'对其进行配置。有关详细信息,请参见:
有人可以向n00bie数据库解释lazy='dynamic'的含义和作用吗?类似地,如果backref存在相同的问题,则需要将backref='items'替换为backref=db.backref'items',lazy='dynamic'。如果没有lazy='dynamic',则直接在env.environment\u资源上获得结果。但是使用lazy='dynamic',它会返回一个-youcando-streng.environment\u资源,它会给你一个SQL查询,你可以在上面加上过滤器等等
_resources = relationship("EnvironmentResource",
cascade="all, delete-orphan",
lazy='dynamic',
passive_deletes=True)