Is there a way to perform a Metadata.create_all() but have it only create the tables, without any of the FKs? And then create the FKs in one go after the fixture data has been loaded into the DB?
On Friday, December 11, 2015 at 3:38:16 PM UTC-8, Michael Bayer wrote: > > > > On 12/11/2015 05:25 PM, Gerald Thibault wrote: > > I am basing my question off the code > > at > http://docs.sqlalchemy.org/en/latest/orm/relationship_persistence.html#rows-that-point-to-themselves-mutually-dependent-rows, > > > > with a few changes. > > > > I am trying to handle a situation very similar to the one in that > > example, with 2 classes having the same relationship types as those in > > the example. However, I would like to create the instances without using > > the relationships, and instead populate the fk values directly. The > > example uses this, and it works. > > > > w1 = Widget(name='somewidget') > > e1 = Entry(name='someentry') > > w1.favorite_entry = e1 > > w1.entries = [e1] > > session.add_all([w1, e1]) > > session.commit() > > > > I would like to do this: > > > > w1 = Widget(widget_id=1, favorite_entry_id=1, name='somewidget') > > e1 = Entry(entry_id=1, widget_id=1, name='someentry') > > session.add_all([w1, e1]) > > session.commit() > > > > The reason I am doing it this way is because I am operating from a JSON > > fixture file, and trying to populate a database for unit testing. The > > method used in the example works perfectly, but trying to do it my way > > yields: > > > > sqlalchemy.exc.IntegrityError: (IntegrityError) (1452, 'Cannot add or > > update a child row: a foreign key constraint fails (`test`.`widget`, > > CONSTRAINT `fk_favorite_entry` FOREIGN KEY (`favorite_entry_id`) > > REFERENCES `entry` (`entry_id`))') 'INSERT INTO widget (widget_id, > > favorite_entry_id, name) VALUES (%s, %s, %s)' (1, 1, 'somewidget') > > > > I understand the post_update option is on the relationship, and not the > > column, so it has no effect on column population. Is there an > > alternative method to have that column populated separately via a second > > statement, similar to the post_update functionality? > > Sure, you have to do it manually: > > w1 = Widget(widget_id=1, favorite_entry_id=None, name='somewidget') > e1 = Entry(entry_id=1, widget_id=1, name='someentry') > session.add_all([w1, e1]) > session.flush() # optional if you're on autoflush > session.query(Widget).filter_by(widget_id=1).update(favorite_entry_id=1) > session.commit() > > > > > > > > > > > -- > > You received this message because you are subscribed to the Google > > Groups "sqlalchemy" group. > > To unsubscribe from this group and stop receiving emails from it, send > > an email to sqlalchemy+...@googlegroups.com <javascript:> > > <mailto:sqlalchemy+unsubscr...@googlegroups.com <javascript:>>. > > To post to this group, send email to sqlal...@googlegroups.com > <javascript:> > > <mailto:sqlal...@googlegroups.com <javascript:>>. > > Visit this group at http://groups.google.com/group/sqlalchemy. > > For more options, visit https://groups.google.com/d/optout. > -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/d/optout.
