On Jan 14, 2008, at 12:44 PM, Alexandre Conrad wrote:
> > Michael Bayer wrote: >> your best bet with this mapping right now is: >> >> print >> Media >> .query >> .select_from >> (media_table >> .join >> (catalog_table >> ).join >> (catalog_channel_table >> )).filter(CatalogChannel.c.id_channel==playlist.id_channel).all() >> >> which is really how select_from() was intended to be used. > > This works with SQLite, but not MySQL: (1054, "Unknown column > 'catalog_channels.id_channel' in 'on clause'") > that doesnt sound right. taking out select_table, and doing: print Media .query .select_from (media_table .join (catalog_table ).join (catalog_channel_table )).filter(CatalogChannel.c.id_channel==playlist.id_channel).all() leads to the SQL: SELECT medias.id AS medias_id, medias.name AS medias_name, medias.id_catalog AS medias_id_catalog FROM medias JOIN catalogs ON catalogs.id = medias.id_catalog JOIN catalog_channels ON catalogs.id = catalog_channels.id WHERE catalog_channels.id_channel = ? ORDER BY medias.oid which is entirely acceptable (and works in mysql). --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To post to this group, send email to [email protected] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~----------~----~----~----~------~----~------~--~---
