In case anyone was going to attempt to solve this, I have come up with a better way to approach the problem. The data is still structured the same way, but I am now going to traverse it in a different way, such that I no longer need a Foo to be aware of everywhere it is being referenced from.
Still, if anyone is aware of a nice solution to the problem I originally presented, please, feel free to share it! Cheers Cameron Jackson Engineering Intern Air Operations Thales Australia Thales Australia Centre, WTC Northbank Wharf, Concourse Level, Siddeley Street, Melbourne, VIC 3005, Australia Tel: +61 3 8630 4591 cameron.jack...@thalesgroup.com.au<mailto:cameron.jack...@thalesgroup.com.au> | www.thalesgroup.com.au<http://www.thalesgroup.com.au> From: sqlalchemy@googlegroups.com [mailto:sqlalchemy@googlegroups.com] On Behalf Of Jackson, Cameron Sent: Wednesday, 4 January 2012 10:41 AM To: sqlalchemy@googlegroups.com Subject: [sqlalchemy] Is there a way to accrue all references to a declarative mapped object? I have a relatively complex structure to my data. In essence it's something like: * A Foo is a small object containing some data. * A Bar has a fixed number of (say, 3) references to Foos * A Baz has a single Bar, plus one additional Foo * A Qux has a single Foo I know it seems like an odd structure, but there is other data present at each level, and you're just going to have to take my word for it that it does make sense! :P Here are the SQLA declarative classes representing the above: class Foo(Base): __tablename__ = 'foos' id = Column(Integer, primary_key = True) name = Column(String) class Bar(Base): __tablename__ = 'bars' id = Column(Integer, primary_key = True) primary_foo_id = Column(Integer, ForeignKey('foos.id')) secondary_foo_id = Column(Integer, ForeignKey('foos.id')) tertiary_foo_id = Column(Integer, ForeignKey('foos.id')) primary_foo = relationship(Foo, primaryjoin = (primary_foo_id == Foo.id)) secondary_foo = relationship(Foo, primaryjoin = (secondary_foo_id == Foo.id)) tertiary_foo = relationship(Foo, primaryjoin = (tertiary_foo_id == Foo.id)) class Baz(Base): __tablename__ = 'quxs' id = Column(Integer, primary_key = True) bar_id = Column(Integer, ForeignKey('bars.id')) extra_foo_id = Column(Integer, ForeignKey('foos.id')) bar = relationship(Bar) extra_foo = relationship(Foo) class Qux(Base): __tablename__ = quxs id = Column(Integer, primary_key = True) foo_id = Column(Integer, ForeignKey('foos.id')) foo = relationship(Foo) I should note that there is no rule that says a Bar cannot use the same Foo twice, e.g. as both its secondary and tertiary Foos. Likewise, a Baz's extra Foo could be the same as one of the Foos of its Bar. Now, to my actual problem. Given a Foo object, I want an easy way to gather all references to it. So I want a list of every Bar, Baz and Qux that references the Foo object, and if a Bar references the Foo two or three times, it should appear two or three times in the list. So far the only solution I've come up with is to put a uniquely named backref into each of the above relationships (they would be somethinig like 'bars_as_primary', 'bars_as_secondary', 'bars_as_tertiary', 'bazs', and 'quxs'), and then add a method, Foo.GetUsages(), which chains all of these arrays together and returns the result. That seems a bit verbose, but then it's not exactly a textbook problem. There's no magical function that returns a list of all of the objects that have a reference to some specific other object is there? If you've read this far, thanks! Cheers, Cam Cameron Jackson Engineering Intern Air Operations Thales Australia Thales Australia Centre, WTC Northbank Wharf, Concourse Level, Siddeley Street, Melbourne, VIC 3005, Australia Tel: +61 3 8630 4591 cameron.jack...@thalesgroup.com.au<mailto:cameron.jack...@thalesgroup.com.au> | www.thalesgroup.com.au<http://www.thalesgroup.com.au> ------------------------------------------------------------------------- DISCLAIMER: This e-mail transmission and any documents, files and previous e-mail messages attached to it are private and confidential. They may contain proprietary or copyright material or information that is subject to legal professional privilege. They are for the use of the intended recipient only. Any unauthorised viewing, use, disclosure, copying, alteration, storage or distribution of, or reliance on, this message is strictly prohibited. No part may be reproduced, adapted or transmitted without the written permission of the owner. 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