Ich bin ziemlich neu in SQLAlchemy oder sogar Datenbank-Programmierung, vielleicht ist meine Frage zu einfach. Jetzt habe ich zwei Klassen/Tabelle:Starter Frage des deklarativen Stils SQLAlchemy relation()
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))
Natürlich kann es nicht laufen. Dies ist der Backtrace:
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.
Es gibt zwei Fremdschlüssel in der Klasse Computer, so dass die Beziehung() callings nicht bestimmen kann, welche verwendet werden soll. Ich denke, ich muss zusätzliche Argumente verwenden, um es zu spezifizieren, richtig? Und wie? Dank
Dank für Ihre Beratung. Der Tippfehler wurde behoben und Backtrace wurde angehängt. – jfding
Danke, Problem gelöst. – jfding