Thanks that worked beautifully.
On a similar note, how would I match documents with only the tags that
I specify in the list? My naive attempt is:
for tag in tag_list:
session.query(Document).join(Document.tags).filter_by(tag=tag)
But that doesn't work.
On Mar 15, 10:54 am, "Michael Bayer" <mike...@zzzcomputing.com> wrote:
> Stodgewrote:
> > 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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