Ok so far I have this:
expressions = []
for tag in tag_list:
expressions += session.query(Document).filter(Tag.tag==tag)
documents =
session.query(Document).join(Document.tags).filter(and_(*expressions))
Doesn't work but it's progress! :)
On Mar 18, 2:37 pm, Stodge <sto...@gmail.com> wrote:
> 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]))
>
> > > --
> > > You received this message because you are subscribed to the Google Groups
> > > "sqlalchemy" group.
> > > To post to this group, send email to sqlalch...@googlegroups.com.
> > > To unsubscribe from this group, send email to
> > > sqlalchemy+unsubscr...@googlegroups.com.
> > > For more options, visit this group at
> > >http://groups.google.com/group/sqlalchemy?hl=en.
>
>
--
You received this message because you are subscribed to the Google Groups
"sqlalchemy" group.
To post to this group, send email to sqlalch...@googlegroups.com.
To unsubscribe from this group, send email to
sqlalchemy+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/sqlalchemy?hl=en.
