sess.query(Website).filter(Website.materials.any())
neat huh ?
that "materials=[]" declaration is not needed BTW, the collection
defaults to an empty list due to the relation() set up on it.
On Sep 1, 2008, at 1:50 PM, Gustavo Narea wrote:
>
> Hello, everybody.
>
> In my model, I have a websites table and another table for the
> badges for
> every website; where some web sites may not have a badge at all.
>
> So, I'm looking for a query which returns the websites that have at
> least one
> badge; it doesn't matter how many badges they have, only _if_ they
> have a
> badge.
>
> This is how my model looks like:
>
>> class Website(DeclarativeEntity):
>> __tablename__ = 'website'
>> website_alias = Column(Unicode(15), primary_key=True)
>> materials = []
>>
>>
>> class Badge(DeclarativeEntity):
>> __tablename__ = badge
>> _website_alias = Column('website_alias', Unicode(15),
>> ForeignKey('website.website_alias'))
>> badge_name = Column(Unicode(20), primary_key=True)
>> website = relation('Website', backref='materials')
>
> Thanks in advance.
> --
> Gustavo Narea.
> General Secretary. GNU/Linux Matters.
> http://www.gnulinuxmatters.org/
>
> >
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