Inserting new records with one-to-many relationship in sqlalchemy

Question:

I’m following the flask-sqlalchemy tutorial on declaring models regarding one-to-many relationship. The example code is as follows:

class Person(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(50))
    addresses = db.relationship('Address', backref='person',
                                lazy='dynamic')

class Address(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    email = db.Column(db.String(50))
    person_id = db.Column(db.Integer, db.ForeignKey('person.id'))

Now I’m wondering how to insert new records into DB using such model. I assume I need a constructor init, but I have difficulties to understand how it should be implemented and used. The main problem for me here is that Person depends on Address and Address has ForeignKey to Person, so it should know about the Person in advance.

Plase help me to understand how it should be performed.

Thank you in advance.

Asked By: wanderlust

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Source

Answers:

You dont need to write a constructor, you can either treat the addresses property on a Person instance as a list:

a = Address(email='[email protected]')
p = Person(name='foo')
p.addresses.append(a)

Or you can pass a list of addresses to the Person constructor

a = Address(email='[email protected]')
p = Person(name='foo', addresses=[a])

In either case you can then access the addresses on your Person instance like so:

db.session.add(p)
db.session.add(a)
db.session.commit()
print(p.addresses.count()) # 1
print(p.addresses[0]) # <Address object at 0x10c098ed0>
print(p.addresses.filter_by(email='[email protected]').count()) # 1
Answered By: DazWorrall

The most important thing while looking into this model is to understand the fact that this model has a one to many relationship, i.e. one Person has more than one address and we will store those addresses in a list in our case.

So, the Person class with its init will look something like this.

class Person(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(50))
    addresses = db.relationship('Address', backref='person',
                            lazy='dynamic')

    def __init__(self,id,name,addresses = tuple()):
        self.id = id
        self.name = name
        self.addresses = addresses

So this Person class will be expecting an id, a name and a list that contains objects of type Address. I have kept that the default value to be an empty list.

Hope it helps. 🙂

Answered By: padfoot27

I’ve gathered information here and elsewhere and found 3 ways to do so. In this model example (same as question):

class Person(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(50))
    addresses = db.relationship('Address', backref='person',
                                lazy='dynamic')

class Address(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    email = db.Column(db.String(50))
    person_id = db.Column(db.Integer, db.ForeignKey('person.id'))

1.

a = Address(email='[email protected]')
p = Person(name='foo', addresses=[a])

2.

p = Person(name='foo')
a = Address(email='[email protected]', person_id=p.id)

3.

a = Address(email='[email protected]')
p = Person(name='foo')
p.addresses.append(a)
Answered By: SIMMORSAL

Additionally to all previous answers, in one-to-one relationships with uselist=False, like:

class Person(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(50))
    address = db.relationship('Address', backref='person',
                                lazy=True, uselist=False)

class Address(db.Model):
    id = db.Column(db.Integer, primary_key=True)
    email = db.Column(db.String(50))
    person_id = db.Column(db.Integer, db.ForeignKey('person.id'))

Only next approach helped to insert records:

p = Person(name='foo')
a = Address(email='[email protected]', person=p)  # Adding created person directly to Address's backref
Answered By: D4rkof