[sqlalchemy] Re: Dynamically building a query with a join
there is append_from() joins know how to find their components that are already in the selectable and replace them. On Mar 5, 2007, at 4:38 PM, Dennis wrote: I'm playing around with dynamically building a query. I can append columns, where clauses, from objects etc... but what about the case where I want to modify the from obj with a join? For example I can do this: sel=select() sel.append_from(a) sel.append_from(b) sel.append_whereclause(a.c.id==b.c.id) That won't work for an outerjoin though. I have a query that works like this now: select ( [...], from_obj=[a.outerjoin(b)] ) but I can't figure out a way to add the outerjoin dynamically. I looked at clause visitors but there doesn't seem like a way to actually modify an existing join. Any thoughts? Thanks Dennis --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: Dynamically building a query with a join
Actually, I'm still having a problem because the primary object is already a join and the next object that I append gets listed twice. Example sel=select([a,b], from_obj=[a.outerjoin(b)]) sel.append( a.outerjoin(c,somecriteriaforthejoin)) str(sel) SELECT ,,, FROM a LEFT OUTER JOIN b ON , a LEFT OUTER JOIN c ON ... sel.execute() SQLError: (ProgrammingError) table name a specified more than once What I really need is: a.outerjoin(b).outerjoin(c) That is what I can't seem to find a way to dynamically generate. Thanks -Dennis On Mar 5, 3:16 pm, Michael Bayer [EMAIL PROTECTED] wrote: there is append_from() joins know how to find their components that are already in the selectable and replace them. On Mar 5, 2007, at 4:38 PM, Dennis wrote: I'm playing around with dynamically building a query. I can append columns, where clauses, from objects etc... but what about the case where I want to modify the from obj with a join? For example I can do this: sel=select() sel.append_from(a) sel.append_from(b) sel.append_whereclause(a.c.id==b.c.id) That won't work for an outerjoin though. I have a query that works like this now: select ( [...], from_obj=[a.outerjoin(b)] ) but I can't figure out a way to add the outerjoin dynamically. I looked at clause visitors but there doesn't seem like a way to actually modify an existing join. Any thoughts? Thanks Dennis --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
[sqlalchemy] Re: Dynamically building a query with a join
I did find a slight hack: query.append_from ( query.froms._list[0].outerjoin( etc ... ) ) -Dennis On Mar 5, 3:54 pm, Dennis [EMAIL PROTECTED] wrote: Actually, I'm still having a problem because the primary object is already a join and the next object that I append gets listed twice. Example sel=select([a,b], from_obj=[a.outerjoin(b)]) sel.append( a.outerjoin(c,somecriteriaforthejoin)) str(sel) SELECT ,,, FROM a LEFT OUTER JOIN b ON , a LEFT OUTER JOIN c ON ... sel.execute() SQLError: (ProgrammingError) table name a specified more than once What I really need is: a.outerjoin(b).outerjoin(c) That is what I can't seem to find a way to dynamically generate. Thanks -Dennis On Mar 5, 3:16 pm, Michael Bayer [EMAIL PROTECTED] wrote: there is append_from() joins know how to find their components that are already in the selectable and replace them. On Mar 5, 2007, at 4:38 PM, Dennis wrote: I'm playing around with dynamically building a query. I can append columns, where clauses, from objects etc... but what about the case where I want to modify the from obj with a join? For example I can do this: sel=select() sel.append_from(a) sel.append_from(b) sel.append_whereclause(a.c.id==b.c.id) That won't work for an outerjoin though. I have a query that works like this now: select ( [...], from_obj=[a.outerjoin(b)] ) but I can't figure out a way to add the outerjoin dynamically. I looked at clause visitors but there doesn't seem like a way to actually modify an existing join. Any thoughts? Thanks Dennis --~--~-~--~~~---~--~~ 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 [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en -~--~~~~--~~--~--~---
