[sqlalchemy] Re: Advice on complicated (?) SQL query

Thanks! Based on your suggestion, I tried the following: The inner query goes like this: repetition_table_2 = repetition_table.alias() s_inner = select([repetition_table_2.c.card_key], (repetition_table_2.c.rep_number==5) \ (repetition_table_2.c.grade==2)).limit(10)

[sqlalchemy] Re: Advice on complicated (?) SQL query

Try this for the outer query: s = select([repetition_table.c.grade],(repetition_table.c.rep_number==2) (repetition_table.c.card_key.in_(s_inner)) ) Barry - Original Message From: pbienst [EMAIL PROTECTED] To: sqlalchemy sqlalchemy@googlegroups.com Sent: Saturday, October 20, 2007

[sqlalchemy] Re: Advice on complicated (?) SQL query

Barry Hart wrote: Try this for the outer query: s = select([repetition_table.c.grade],(repetition_table.c.rep_number==2) (repetition_table.c.card_key.in_(s_inner)) ) Great, that did the trick! Thanks, Peter --~--~-~--~~~---~--~~ You received this