Pregunta inicial del estilo declarativo SQLAlchemy relation()

Question

Soy bastante nuevo en SQLAlchemy, o incluso en la programación de bases de datos, tal vez mi pregunta es demasiado simple. Ahora tengo dos clases/tabla:

class User(Base): 
    __tablename__ = 'users' 
    id = Column(Integer, primary_key=True) 
    name = Column(String(40)) 
    ... 

class Computer(Base): 
    __tablename__ = 'comps' 
    id = Column(Integer, primary_key=True) 
    buyer_id = Column(None, ForeignKey('users.id')) 
    user_id = Column(None, ForeignKey('users.id')) 
    buyer = relation(User, backref=backref('buys', order_by=id)) 
    user = relation(User, backref=backref('usings', order_by=id)) 

Por supuesto, no se puede ejecutar. Este es el trazado inverso:

File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/state.py", line 71, in initialize_instance 
    fn(self, instance, args, kwargs) 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/mapper.py", line 1829, in _event_on_init 
    instrumenting_mapper.compile() 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/mapper.py", line 687, in compile 
    mapper._post_configure_properties() 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/mapper.py", line 716, in _post_configure_properties 
    prop.init() 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/interfaces.py", line 408, in init 
    self.do_init() 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/properties.py", line 716, in do_init 
    self._determine_joins() 
    File "/Library/Python/2.6/site-packages/SQLAlchemy-0.5.8-py2.6.egg/sqlalchemy/orm/properties.py", line 806, in _determine_joins 
    "many-to-many relation, 'secondaryjoin' is needed as well." % (self)) 
sqlalchemy.exc.ArgumentError: Could not determine join condition between parent/child tables on relation Package.maintainer. Specify a 'primaryjoin' expression. If this is a many-to-many relation, 'secondaryjoin' is needed as well. 

Hay dos claves externas de Clase del ordenador, por lo que la relación (llamados) no se puede determinar cuál debe ser utilizado. Creo que debo usar argumentos adicionales para especificarlo, ¿verdad? ¿Y cómo? Gracias

Answer 1

La sintaxis correcta debe ser:

buyer = relation(User, backref=backref('buys', order_by=id)) 
user = relation(User, backref=backref('usings', order_by=id)) 

P. S. La próxima vez, especifique a qué se refiere con "no se puede ejecutar" publicando un rastreo.

actualización: el rastreo en cuestión actualizada dice exactamente lo que necesita: especificar primaryjoin condición:

buyer = relation(User, primaryjoin=(buyer_id==User.id), 
       backref=backref('buys', order_by=id)) 
user = relation(User, primaryjoin=(user_id==User.id), 
       backref=backref('usings', order_by=id))