Hello, and sorry if this issue is already covered.
I have a table which makes foreign references to other table, and I need to
make a filter against that table multiple times.
so I have:
q1=session.query(Fuente).outerjoin(Fuente.PPersona,
aliased=True).outerjoin(Fuente.PPersona_como_fuente, aliased=True)
i.e. Fuente makes more than one reference to table Persona, thru the
properties PPersona and PPersona_como_fuente.
until here, it works fine.
Now if I need to construct a filter against Persona, how do I make explicit
that I want to make a condition to a field of Persona
as, for example, Fuente.PPersona.sexo and not
Fuente.PPersona_como_fuente.sexo ?
lets say:
q2=q1.filter(Persona.sexo == '111')
I get an ambiguous sql expression, like this:
SELECT
fuente.id AS fuente_id, fuente.caso_id AS fuente_caso_id
FROM fuente
LEFT OUTER JOIN persona AS persona_1 ON persona_1.id =
fuente.persona_referenciada_id
LEFT OUTER JOIN persona AS persona_2 ON persona_2.id =
fuente.persona_fuente_id
WHERE
persona_2.sexo = %(persona_sexo_1)s
ORDER BY fuente.id
Until now, the only solution I found was to make an alias to the 'persona'
table, and then write new definitions for alternate 'Persona' object class
and PersonaMapper.
That way it works fine, but if I work with that table from a complex set of
references, it becomes a nightmare of copies....
BTW, in spanish, sexo (also) means gender ;)
So thanks in advance for any magic solution (or reference)
Adolfo
--
Adolfo Dunayevich
--~--~---------~--~----~------------~-------~--~----~
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
-~----------~----~----~----~------~----~------~--~---
