Kirk Strauser wrote: > > invoices = session.query(Invoice) > invoices = invoices.join(BillingInfo) > invoices = invoices.join((Customer, > BillingInfo.xrscustid==Customer.xrscustid)) > > invoices = invoices.filter(BillingInfo.typeship=='GBL') > invoices = invoices.filter(Invoice.invid==2663703)
at this point, invoices[0] is the invoice that is subject to the given filter criterion. > > print invoices[0].BillingInfo.typeship now this part is very unusual and is something I haven't tested. Your foreign key is to only one column of a composite primary key - very strange. So SQLA probably sees this as "many-to-one" but the result in which the lazy load will incur is essentially random since many BillingInfo entries may have that same value. Its very likely that your linkage here is incorrect, and you in fact want to declare, at least within SQLAlchemy-land, a composite foreign key (using ForeignKeyConstraint) on Invoice that matches both invoice.xrscust and invoice.pay2addrid to both of the corresponding columns on BillingInfo. The explicit join condition on the relation() would then no longer be needed (the need to explicitly declare things SQLA should be figuring out may be considered a code smell here). --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To post to this group, send email to sqlalchemy@googlegroups.com To unsubscribe from this group, send email to sqlalchemy+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~----------~----~----~----~------~----~------~--~---
