Thanks. I wrote the error message from memory, the exact wording is:
session.query(Document).filter(Document.tags.in_(tag_list))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python2.6/site-packages/sqlalchemy/sql/
expression.py", line 1274, in in_
return self.operate(operators.in_op, other)
File "/usr/lib/python2.6/site-packages/sqlalchemy/orm/
attributes.py", line 120, in operate
return op(self.comparator, *other, **kwargs)
File "/usr/lib/python2.6/site-packages/sqlalchemy/sql/operators.py",
line 49, in in_op
return a.in_(b)
File "/usr/lib/python2.6/site-packages/sqlalchemy/orm/
properties.py", line 479, in in_
raise NotImplementedError("in_() not yet supported for relations.
For a "
NotImplementedError: in_() not yet supported for relations. For a
simple many-to-one, use in_() against the set of foreign key values.
Thanks for the suggestions.
On Mar 15, 10:54 am, "Michael Bayer" <[email protected]> wrote:
> Stodge wrote:
> > I have two classes with a third table:
>
> > document_tags = Table('document_tags', metadata,
> > Column('document_id', Integer, ForeignKey('documents.id')),
> > Column('tag_id', Integer, ForeignKey('tags.id'))
> > )
>
> > class Document(Base):
> > __tablename__ = 'documents'
>
> > id = Column(Integer, primary_key=True)
> > title = Column(String)
> > filename = Column(String)
> > tags = relation('Tag', secondary=document_tags, backref='tags')
>
> > def __init__(self, title, filename):
> > self.title = title
> > self.filename = filename
>
> > class Tag(Base):
> > __tablename__ = 'tags'
>
> > id = Column(Integer, primary_key=True)
> > tag = Column(String)
>
> > def __init__(self, tag):
> > self.tag = tag
>
> > I want to find all documents with tags in a given list of tags:
>
> > documents =
> > session.query(Document).filter(Document.tags.in_(tag_list))
>
> > except I get the familiar message that the "in_()" operator is not
> > currently implemented for many-to-one-relations.
>
> > I've searched and found some alternatives but I can't get any to work.
> > Is there an easy example that will make this work? Thanks
>
> if the error message says "many-to-one" then that's a bug. Your relation
> is many-to-many.
>
> in this case the syntactically easiest method is to use any().
> Document.tags.any(Tag.id.in_([t.id for t in tag_list])).
>
> A join could be more performant, which would be:
>
> query.join(Document.tags).filter(Tag.id.in_([t.id for t in tag_list]))
>
>
>
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>
>
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