sqlalchemy中hybrid_property中的嵌套关​​系

Question

I have a @hybrid_property which references a nested relationship self.establishment_type.establishment_base_type.name == 'QSR' . It works on a Location object as in assert location.is_qsr == True , but not in a filter. I have tried adding a @is_qsr.expression function, but can't get any of them working. How can I enable a filter such as query(Location).filter(Location.is_qsr == True) ?

class Location(Base):
    __tablename__ = 'houses'
    id = Column(Integer, primary_key=True)
    establishment_type_id = Column(
        Integer, ForeignKey('establishment_types.id')
    )
    establishment_type = relationship('EstablishmentType')

    @hybrid_property
    def is_qsr(self):
        if self.establishment_type:
            if self.establishment_type.establishment_base_type:
                return self.establishment_type.establishment_base_type.name == 'QSR'

        return False

class EstablishmentType(Base):
    __tablename__ = 'establishment_types'
    id = Column(Integer, primary_key=True)

    establishment_base_type_id = Column(
        Integer, ForeignKey('establishment_base_types.id')
    )
    establishment_base_type = relationship('EstablishmentBaseType')

class EstablishmentBaseType(Base):
    __tablename__ = 'establishment_base_types'
    id = Column(Integer, primary_key=True)

Answer 1

You can use .has on relationships:

@is_qsr.expression
def is_qsr(cls):
    return cls.establishment_type.has(
        EstablishmentType.establishment_base_type.has(
            EstablishmentBaseType.name == "QSR"))

This doesn't produce the most efficient query in the world (it does a EXISTS (SELECT 1 FROM ...) ) but a decent optimizer should be able to figure it out.