Hi I use 0.8's column_expression. Like this:
----- from sqlalchemy.types import UserDefinedType from sqlalchemy.sql import func class Geometry(UserDefinedType): def column_expression(self, col): return func.ST_AsBinary(col, type_=self) from sqlalchemy import Table, Column, MetaData lakes = Table('lake', MetaData(), Column('geom', Geometry) ) from sqlalchemy.sql import select s = select([lakes]) print s ---- The final print statement returns this: "SELECT ST_AsBinary(lake.geom) AS geom_1 FROM lake". My issue is with the "geom_1" label being generated. My column name being "geom" I'd expect the following to work: s = select([lakes]) for row in conn.execute(s): geom = row['geom'] but it won't work because "row" does not have a "geom" item. Is there a solution to this issue? Thanks, -- Eric Lemoine Camptocamp France SAS Savoie Technolac, BP 352 73377 Le Bourget du Lac, Cedex Tel : 00 33 4 79 44 44 96 Mail : eric.lemo...@camptocamp.com http://www.camptocamp.com -- 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.