Just do it in Python: res = [] for hpoint in session.query(HistoryPoint).all(): res.setdefault(hpoint.vehicle_id, []).append({'id':hpoint.id,'location': str(hpoint.location)})
On Tuesday, September 22, 2015 at 6:48:14 PM UTC+3, Johnny W. Santos wrote: > > Supose I have the models below, how could I query for a result like this: > > [ > {1: [{"id": 3, "location": "POINT(23.23423423 54.234524234)"},{"id": 4, > "location": "POINT(23.23423423 54.234524234)"}]}, > {2: [{"id": 45, "location": "POINT(78.23423423 43.234524234)"},{"id": 67 > , "location": "POINT(34.2347683423 74.234524234)"}]}, > ] > > The keys being the vehicle id and a list of HistoryPoints values. > > Supposing I have more fields in HistoryPoint and is prefered to query on > it due some filter conditions. > > > class Vehicle(Base): > id = Column(Integer, primary_key=True) > identifier = Column(Integer) > > > class HistoryPoint(Base): > vehicle_id = Column(Integer, FokeignKey('vehicles.id')) > location = Column(Geography('POINT')) > > vehicle = relationship('Vehicle', backref='history') > > > I'm really having a hard time on this, any help will be appreciated > > Johnny > -- 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.