Sorry, that should have been: 

existing_creator = DBSession(Creator).query.filter_by(creator=creator).first() 


On Monday, August 12, 2013 at 9:49 PM, Tim Van Steenburgh wrote:

> It's not the append that's causing the error, it's the fact that you're 
> creating a new Creator() instance, which ultimately results in an INSERT 
> statement being issued. 
> 
> You want to append a Creator instance to `company.creator`, but you don't 
> necessarily want to make a new Creator every time you instantiate a Company. 
> If a Creator with the given name already exists, you'll want use that instead.
> 
> So, roughly:
> 
> class Company(Base):
>     __tablename__ = "companies"
>     id = Column(Integer, primary_key = True)
>     company = Column(String(100), unique=True, nullable=False)
>     creator = relationship("Creator", backref="companies", cascade="all")
>     def __init__(self, company, creator):
>         self.company = company
>   existing_creator = DBSession(Creator).query.filter_by(name=creator).first()
>         self.creator.append(existing_creator or Creator(creator))
> 
> 
> -- 
> Tim Van Steenburgh
> 
> 
> 
> On Monday, August 12, 2013 at 9:41 PM, [email protected] 
> (mailto:[email protected]) wrote:
> 
> > Sorry I don't understand what you're trying to say. 
> > 
> > If the Creator already exists, and I'm to append it again, isn't that the 
> > same as what my code is currently doing? (That is, appending in every 
> > instance.) I don't see how this wouldn't result in the same error message.
> > 
> > And what would it mean to create a new one and append that? I don't know 
> > what this code would look like.
> > 
> > Apologies if I'm being dense.
> > 
> > On Monday, August 12, 2013 9:33:31 PM UTC-4, Tim wrote:
> > > In `Company.__init__()`, instead of blindly creating a new `Creator` 
> > > instance, you need to first query for an existing Creator with that name. 
> > > If it exists, append it, otherwise, create a new one and append that. 
> > > 
> > > -- 
> > > Tim Van Steenburgh
> > > 
> > > 
> > > On Monday, August 12, 2013 at 9:26 PM, [email protected] (javascript:) 
> > > wrote:
> > > 
> > > > I have another question about a piece of code that I posted the other 
> > > > day. Namely, I have a one-to-many relationship between Creator and 
> > > > Company. A Creator can have a relationship with multiple Companies but 
> > > > any one Company can have a relationship with only one Creator.
> > > > 
> > > > class Company(Base):
> > > >     __tablename__ = "companies"
> > > >     id = Column(Integer, primary_key = True)
> > > >     company = Column(String(100), unique=True, nullable=False)
> > > >     creator = relationship("Creator", backref="companies", 
> > > > cascade="all")
> > > > 
> > > > 
> > > >     def __init__(self, company, creator):
> > > >         self.company = company
> > > >         self.creator.append(Creator(creator))
> > > > 
> > > > class Creator(Base):
> > > >     __tablename__ = "creators"
> > > > 
> > > >     company_id = Column(Integer, ForeignKey('companies.id 
> > > > (http://companies.id/)'))
> > > >     creator = Column(String(100), nullable=False, unique=True)
> > > > 
> > > >     def __init__(self, creator):
> > > >         self.creator = creator
> > > > 
> > > > 
> > > > So, to create a Company, the code calls company = Company(<company 
> > > > name>, <creator name>) and that in turn calls Creator().
> > > > 
> > > > The problem is that the Companies get added one by one, and if a new 
> > > > company being entered has a Creator with a name of a preexisting 
> > > > company, SQLalchemy errors due to the unique=True flag:
> > > > 
> > > > sqlalchemy.exc.IntegrityError: (IntegrityError) (1062, "Duplicate entry 
> > > > 'Viking' for key 'creator'") 'INSERT INTO creators (company_id, 
> > > > creator) VALUES (%s, %s)' (17L, u'Viking')
> > > > 
> > > > If unique=True isn't enabled, it will create another Creator of the 
> > > > same name. Instead, the code should reflect the additional Company 
> > > > assigned to this particular Creator. How might I go about fixing this?
> > > > 
> > > > Thanks! 
> > > > 
> > > > -- 
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> > > >  
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