Hi,
When SQALchemy can't guess the join condition to use for a relationship,
you have to use the primaryjoin parameter :
ursmst = relationship('Usrmst',
primaryjoin="Workgrp.workgrp_owner==Usrmst.id")
Hope this helps.
Regards
Gaston
Le 07/02/2018 à 12:01, Jeremy Flowers a écrit :
> Hi
> I've recently used sqlacodegen
>
> When I try and run against the generated code it get this message
> that I've been unable to fix:
>
> Could not determine join condition between parent/child tables on
> relationship Workgrp.usrmst - there are multiple foreign key paths
> linking the tables. Specify the 'foreign_keys' argument, providing a
> list of those columns which should be counted as containing a foreign
> key reference to the parent table.
>
> I'm wondering if someone can shed some light:
>
> Here is the generated code for the relevant tables causing the issue:
>
> class Workgrp(Owner):
> __tablename__ = 'workgrp'
>
> workgrp_id = Column(ForeignKey('owner.owner_id'), primary_key=True)
> workgrp_prntid = Column(Numeric(scale=0, asdecimal=False))
> workgrp_name = Column(String(256))
> workgrp_desc = Column(String(4000))
> workgrp_owner = Column(ForeignKey('usrmst.usrmst_id'))
> workgrp_lstchgtm = Column(DateTime, index=True)
> workgrp_externid = Column(String(20))
> workgrp_profile = Column(Text)
> workgrp_usrmodtm = Column(DateTime)
>
> usrmst = relationship('Usrmst')
>
>
> class Usrmst(Owner):
> __tablename__ = 'usrmst'
> __table_args__ = (
> Index('usrmst_ak1', 'usrmst_domain', 'usrmst_name'),
> )
>
> usrmst_id = Column(ForeignKey('owner.owner_id'), primary_key=True)
> usrmst_domain = Column(String(256))
> usrmst_name = Column(String(256), nullable=False)
> usrmst_fullname = Column(String(1024))
> usrmst_desc = Column(String(4000))
> usrmst_phoneno = Column(String(40))
> usrmst_pagerno = Column(String(40))
> usrmst_email = Column(String(1024))
> usrmst_emailtype = Column(Numeric(scale=0, asdecimal=False))
> secmst_id = Column(ForeignKey('secmst.secmst_id'))
> lngmst_id = Column(ForeignKey('lngmst.lngmst_id'))
> usrmst_password = Column(String(1024))
> usrmst_externid = Column(String(20))
> usrmst_suser = Column(String(1))
> usrmst_lstchgtm = Column(DateTime, index=True)
> usrmst_orapassword = Column(String(144))
> usrmst_wingroup = Column(String(1))
> usrmst_tmpacct = Column(String(1))
> usrmst_profile = Column(Text)
> usrmst_usrmodtm = Column(DateTime)
> usrmst_principal = Column(String(256))
> usrmst_keytab = Column(String(4000))
>
> lngmst = relationship('Lngmst')
> secmst = relationship('Secmst')
>
> I've looked at the SQLAlchemy docs, and tried things with
> foreign_keys, primaryjoin,
> http://docs.sqlalchemy.org/en/latest/orm/relationship_api.html#sqlalchemy.orm.relationship.params.foreign_keys
> http://docs.sqlalchemy.org/en/latest/orm/join_conditions.html#relationship-custom-foreign
>
> But ended up with messages like '
> Table' object has no attribute 'usrmst_id'
> AttributeError: 'Table' object has no attribute 'workgrp_owner'
>
> Also for Mike Bayer: Why is there no consistency in naming within
> SQLAlchemy?:
> foreign_keys uses underscore
> primaryjoin doesn't
>
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> The Python SQL Toolkit and Object Relational Mapper
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