Hi Thanks for the hint. FYI: Finally i decided to use hybrid_property and conditional logic using this model
class Employee(Base): __tablename__ = 'employee' id = Column(Integer, primary_key=True) name = Column(String(50)) type = Column(String(50)) __mapper_args__ = { 'polymorphic_identity':'employee', 'polymorphic_on':type } @hybrid_property def POLYMORPHICLS(self): return func.concat(self.__name__) @POLYMORPHICLS.expression def POLYMORPHICLS(self): return case([ ( self.type == 'engineer', func.concat(self.__mapper__.polymorphic_map['engineer'].class_.__name__)), ( self.type == 'manager', func.concat(self.__mapper__.polymorphic_map['manager'].class_.__name__)) ], else_ = func.concat(self.__name__) ) So I can do qury like this : session.query(Employee.POLYMORPHICLS).filter(Employee.name=='Employee_Name'').all() OR FILTER session.query(Employee).filter(Employee.name=='Employee_Name',Employee.POLYMORPHICLS=='Manager').all() I hope this is correct Disadvantage: I have to specify all the existing inherited class type in @POLYMORPHICLS.expression Regards G On Tuesday, June 2, 2015 at 6:30:44 PM UTC+2, g wrote: > > Hi all > > Is there a way to get the "classname" of the polymorphic class result in a > single query on the base class ? > > soemthing like this > > session.query(cast(Employee.__mapper__.polymorphic_map[Employee.type].class_.__name__, > String)).filter(Employee.name=='Employee_Name').one() > > > > *Details * > > *=======* > from sqlalchemy import * > from sqlalchemy.orm import * > from sqlalchemy.ext.declarative import declarative_base > from sqlalchemy.orm import aliased > from sqlalchemy.ext.hybrid import hybrid_property > > Base = declarative_base() > > class Employee(Base): > __tablename__ = 'employee' > id = Column(Integer, primary_key=True) > name = Column(String(50)) > type = Column(String(50)) > __mapper_args__ = { > 'polymorphic_identity':'employee', > 'polymorphic_on':type > } > > class Engineer(Employee): > __tablename__ = 'engineer' > id = Column(Integer, ForeignKey('employee.id'), primary_key=True) > engineer_name = Column(String(30)) > > __mapper_args__ = { > 'polymorphic_identity':'engineer', > } > > class Manager(Employee): > __tablename__ = 'manager' > id = Column(Integer, ForeignKey('employee.id'), primary_key=True) > manager_name = Column(String(30)) > > __mapper_args__ = { > 'polymorphic_identity':'manager', > } > e = create_engine('postgresql+psycopg2://uuuu:****@host/test', echo=False) > Base.metadata.create_all(e) > > session = Session(e) > > *WITH THIS DATA* > *===============* > > eng = Engineer(name= 'Employee_Name', engineer_name= 'Engineer_name') > session.add(eng) > session.commit() > > *SIMPLE QUERY * > *==============* > empl = session.query(Employee).filter(Employee.name=='Employee_Name').one() > print empl <__main__.Engineer at 0x6573c50> > > BUT WHAT I WANT IS ONLY THE CLASS NAME OF POLYMORPHIC CLASS SO I TRIED THIS > > > employee = > session.query(cast(Employee.__mapper__.polymorphic_map['employee'].class_.__name__, > > String)).filter(Employee.name=='Employee_Name').one() print(employee) => > (u'Employee',) > > and with > > e = > session.query(cast(Employee.__mapper__.polymorphic_map['engineer'].class_.__name__, > > String)).filter(Employee.name=='Employee_Name').one() > print(e) > > AND finally to generalize i tried: > e = > session.query(cast(Employee.__mapper__.polymorphic_map[Employee.type].class_.__name__, > > String)).filter(Employee.name=='Employee_Name').one() print(e) > > ---------------------------------------------------------------------------KeyError > Traceback (most recent call > last)<ipython-input-31-45f6c96837bf> in <module>() 1 #generalize----> 2 > e = > session.query(cast(Employee.__mapper__.polymorphic_map[Employee.type].class_.__name__, > String)).filter(Employee.name=='EN').one() 3 print(e) > KeyError: <sqlalchemy.orm.attributes.InstrumentedAttribute object at > 0x0654FD80> > > > I think i am doing something logically wrong . > > QUESTION > > ======== > > Is there a way to get the classname of the polymorphic class result in a > single query ? > > I tried with hybrid_property but i had the same error > > > Some hints ? > > > Regards > > G > > > > > > > > > > > -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To unsubscribe from this group and stop receiving emails from it, send an email to sqlalchemy+unsubscr...@googlegroups.com. To post to this group, send email to sqlalchemy@googlegroups.com. Visit this group at http://groups.google.com/group/sqlalchemy. For more options, visit https://groups.google.com/d/optout.