我收到了这个奇怪的错误,我说的很奇怪,因为我改变了一张不相关的表格。
我正在尝试查询我的tDevice
表格,如下所示:
class TDevice(Base):
__tablename__ = 'tDevice'
ixDevice = Column(Integer, primary_key=True)
ixDeviceType = Column(Integer, ForeignKey('tDeviceType.ixDeviceType'), nullable=False)
ixSubStation = Column(Integer, ForeignKey('tSubStation.ixSubStation'), nullable=False)
ixModel = Column(Integer, ForeignKey('tModel.ixModel'), nullable=True)
ixParentDevice = Column(Integer, ForeignKey('tDevice.ixDevice'), nullable=True)
sDeviceName = Column(Unicode(255), nullable=False)#added
children = relationship('TDevice',
backref=backref('parent', remote_side=[ixDevice]))
device_type = relationship('TDeviceType',
backref=backref('devices'))
model = relationship('TModel',
backref=backref('devices'))
sub_station = relationship('TSubStation',
backref=backref('devices'))
这就是我查询它的方式:
Device = DBSession.query(TDevice).filter(TDevice.ixDevice == device_id).one()
执行此行后,我收到错误:
ArgumentError: relationship 'report_type' expects a class or a mapper argument (received: <class 'sqlalchemy.sql.schema.Table'>)
我所做的唯一更改是在tReportTable
中添加report_type关系
现在看起来像这样:
class TReport(Base):
__tablename__ = 'tReport'
ixReport = Column(Integer, primary_key=True)
ixDevice = Column(Integer, ForeignKey('tDevice.ixDevice'), nullable=False)
ixJob = Column(Integer, ForeignKey('tJob.ixJob'), nullable=False)
ixReportType = Column(Integer, ForeignKey('tReportType.ixReportType'), nullable=False) # added
report_type = relationship('tReportType',
uselist=False,
backref=backref('report'))
device = relationship('TDevice',
uselist=False,
backref=backref('report'))
job = relationship('TJob',
uselist=False,
backref=backref('report'))
我还是SqlAlchemy的新手,所以如果我在迭代另一个表,我似乎无法看到如何添加该关系会导致此错误
答案 0 :(得分:40)
对自己并不满意,因为这是一个如此愚蠢的错误,但这是我的罪魁祸首:
report_type = relationship('tReportType',
uselist=False,
backref=backref('report'))
应该是:
report_type = relationship('TReportType',
uselist=False,
backref=backref('report'))
大写T代替t,我应该引用类,而不是我的实际表名:'tReportType'
- &gt; 'TReportType'