SQLAlchemy Join two tables in (1 to 0..1 ) relationship - python-2.7

Question

I have these two tables in (1 to 0..1) relationship:
models.py
# --------------------------------------------------
class Person(db.Model):
__tablename__ = 'person'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String, nullable=False)
address = db.Column(db.String, nullable=False)
# --------------------------------------------------
class Gift(db.Model):
__tablename__ = 'gift'
id = db.Column(db.Integer, primary_key=True)
person_id = db.Column(db.Integer, db.ForeignKey('person.id'))
person = relationship("Person", uselist=False, backref="gift")
gift_idea = db.Column(db.String, nullable=False)
# --------------------------------------------------
Questions:
Is the structure of my one to one tables correct? I know that in the SQLAlchemy documentation, the relationship code line (person = relationship...) is in the parent class but I need it to be in the child class, is that OK?!
My goal is to display (gift_idea from Gift table) and (person_name, person_address from Person table), if there's a gift_idea for that person. In my search I found people use query.join to get that and I have tried many codes but this following statement at least get me half the content I
need:
views.py
a = session.query(Gift).join(Person)
Could you please assist me with my questions? Thank you for any effort.

Answer 1

As per sqlalchemy One to One relationship pattern definition ...
One To One is essentially a bidirectional relationship with a scalar
attribute on both sides. To achieve this, the uselist flag indicates
the placement of a scalar attribute instead of a collection on the
“many” side of the relationship
In your case you've disabled "uselist" for Gifts by giving "uselist=False" (Actually by default it gets disabled). But you didn't disabled "uselist" for Persons. because of this Person can still hold list of Gifts (One person - Many Gifts).
You can get one to one mapping in two ways here..
1: By using backref function which provides arguments for the reverse side. keep your person as it is.
class Gift(db.Model):
__tablename__ = 'gift'
id = db.Column(db.Integer, primary_key=True)
person_id = db.Column(db.Integer, db.ForeignKey('person.id'))
person = relationship("Person", backref=backref("gift", uselist=False))
gift_idea = db.Column(db.String, nullable=False)
2: Adding scalar attribute on both the sides instead of using backref property.
class Gift(db.Model):
__tablename__ = 'gift'
id = db.Column(db.Integer, primary_key=True)
person_id = db.Column(db.Integer, db.ForeignKey('person.id'))
gift_idea = db.Column(db.String, nullable=False)
person = relationship("Person")
# -----------------------------------------------
class Person(db.Model):
__tablename__ = 'person'
id = db.Column(db.Integer, primary_key=True)
name = db.Column(db.String, nullable=False)
address = db.Column(db.String, nullable=False)
gift = relationship("Gift", uselist=False)
Solution for Question-2:
#val: Gave you straightforward solution for this.. you could also try using lazy loading instead of clearly writing everything over orm query.
list_of_gift_object = session.query(Gift).join(Person).all()
may be you can iterate over the list and use Gift person attribute to get associated person data.
for gift in list_of_gift_object:
print gift.gift_idea, gift.person.name, gift.person.address

Answer 2

In order to define proper 1-to-[0..1] relationship from the child side, you must make sure that the backref has uselist=False set, while the uselist from child to parent will be False by default:
person = db.relationship(
"Person",
backref=db.backref("gift", uselist=False),
)
Then the query is very straightforward to do:
a = (session
.query(
Gift.gift_idea,
Person.name.label("person_name"),
Person.address.label("person_address"),
)
# .select_from(Gift) # #note: optional, but will need it if the first column in the `.query(...)` above will be from table `Person`
.join(Person)
)