given the object `source`, these both work cols = [c.key for c in list(source.__table__.columns)] cols = [c.name for c in sqlalchemy.orm.class_mapper(source.__class__).mapped_table.c]
I'm sure there are other ways. is there an ideal / canonical way of getting this data? for this particular instance, my biggest concern is speed. -- 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.