SQLAlchemy的鄰接表關係屬性

Question

我有一個不同類型的孩子繼承了頂級型號聲明

class HasId(object): 

    @declared_attr 
    def id(cls): 
     return Column('id', Integer, Sequence('test_id_seq'), primary_key=True) 
    ... 
    @declared_attr 
     def triggered_by_id(cls): 
      return Column(Integer, ForeignKey('tests.id'), nullable=True) 

    @declared_attr 
     def triggered(cls): 
      return relationship('TestParent', 
           foreign_keys='TestParent.triggered_by_id', 
           lazy='joined', 
           cascade='save-update, merge, delete, delete-orphan', 
           backref=backref('triggered_by', remote_side=[id]) 
           ) 


class TestParent(HasId, Model): 
    __tablename__ = 'tests' 

    discriminator = Column(String(50)) 

    __mapper_args__ = {'polymorphic_on': discriminator} 


class FooTest(TestParent): 
    __tablename__ = 'footests' 
    __mapper_args__ = {'polymorphic_identity': 'footests'} 
    id = Column(Integer, ForeignKey('tests.id'), primary_key=True) 

    bar = Column(Boolean) 
    ... 

當IO嘗試建立這個數據庫，我出現因我是如何定義的錯誤關於triggered_by關係的backref的remote_side。

完整的錯誤是

ArgumentError: Column-based expression object expected for argument 'remote_side'; got: '<built-in function id>', type <type 'builtin_function_or_method'> 

Answer 1

的解決方案是將backref定義從

backref('triggered', remote_side=[id]) 

改變

backref('triggered', remote_side='TestParent.id')