Hi Michael and all....
I successfully built a "all nodes" relationship following your guidelines.
As:
subq1=select([caso_vinculo.c.caso_1_id.label('id1'),caso_vinculo.c.caso_2_id.label('id2')]).union(select([caso_vinculo.c.caso_2_id,
caso_vinculo.c.caso_1_id]))
subq2=aliased(subq1)
CasoMapper = mapper(Caso, caso, properties=
{
'LinkedNodes':relation(Caso, secondary=subq2,
primaryjoin=caso.c.id ==
subq2.c.id1,
secondaryjoin=subq2.c.id2==caso.c.id),
So with this property I can get both left and right linked nodes.
with:
CasoAlias=aliased(Caso)
q1=session.query(Caso.id).outerjoin(CasoAlias.LinkedNodes)
q1.all()
works fine
and once I add a new column from the aliased entity (CasoAlias)
q2=q1.add_column(CasoAlias.id)
it compiles fine
SELECT caso.id AS caso_id, caso_1.id AS caso_1_id
FROM caso AS caso_1 LEFT OUTER JOIN (SELECT caso_vinculo.caso_1_id AS id1,
caso_vinculo.caso_2_id AS id2
FROM caso_vinculo UNION SELECT caso_vinculo.caso_2_id AS caso_2_id,
caso_vinculo.caso_1_id AS caso_1_id
FROM caso_vinculo) AS anon_1 ON caso_1.id = anon_1.id1 LEFT OUTER JOIN caso
ON anon_1.id2 = caso.id
BUT
q2.all()
breaks with
Traceback (most recent call last):
File "<pyshell#6>", line 1, in <module>
q2.all()
File "build\bdist.win32\egg\sqlalchemy\orm\query.py", line 2115, in all
return list(self)
File "build\bdist.win32\egg\sqlalchemy\orm\query.py", line 2351, in
instances
labels) for row in fetch]
File "build\bdist.win32\egg\sqlalchemy\orm\query.py", line 3342, in proc
return row[column]
File "build\bdist.win32\egg\sqlalchemy\engine\base.py", line 2616, in
__getitem__
processor, obj, index = self._parent._key_fallback(key)
File "build\bdist.win32\egg\sqlalchemy\engine\base.py", line 2835, in
_key_fallback
expression._string_or_unprintable(key))
NoSuchColumnError: "Could not locate column in row for column '%(34374288
caso)s.id'"
I am using sa '0.7.8'
Any lead on this error?
Thanks in advance for any help!!
Adolfo
On Monday, August 20, 2012 5:07:39 PM UTC-5, adolfo wrote:
>
> I have a Self-Referential Many-to-Many Relationship situation where the
>
> right_nodes = relationship("Node",
> secondary=node_to_node,
> primaryjoin=id==node_to_node.c.left_node_id,
> secondaryjoin=id==node_to_node.c.right_node_id,
> backref="left_nodes"
>
>
> works fine.
> The problem:
> I need a "related nodes" relationship, which, in one expression, returns
> all related nodes, both left nodes and right nodes, excluding the given
> node itself.
> Is that possible using the RELATIONSHIP construct?
>
> Adolfo
>
>
>
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