On Apr 16, 2009, at 2:31 PM, dykang wrote:
> inside = table.alias('inside') > > outside = inside.join(table, table.c.foo < 2) > s = outside.select(use_labels=True).alias('outside') > > j = table2.join(s, table2.c.bar == outside.c.inside_foo) > > s = j.select(use_labels=True) > rows = session.query(Test2).add_entity(Test, alias=inside).add_entity > (Test, alias=outside).from_statement(s).all() when you create a selectable, the columns on that selectable (i.e. the '.c.' namespace) are children of that selectable. so if you had two selectables: SELECT id, name FROM table and SELECT id, name FROM (select id, name FROM table) both have "id" and "name" in their column list, but have a different parent selectable. you could have instead said: SELECT "something" AS id, name FROM (select id, name FROM table) and its more apparent above how "id" means something different based on which selectable it belongs to. So above, you're joining "table2" to "s". the ON clause must be in terms of "table2" and "s", not "outside", which is meaningless in that context. --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---