python - 重写 sqlalchemy 中的关系行为

Question

假设我以声明方式拥有三个表，Parent、Child 和 Pet，这样

• Parent 与 Child 和 Pet 都是多对多关系
• Child 与 Pet
是一对多关系

他们的代码是(使用 Flask-SQLAlchemy，尽管我相信解决方案存在于 SQLAlchemy 领域而不是 Flask 中)。

class Parent(db.Model):
    __tablename__ = 'parents'

    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(64))

    # many to many relationship between parent and children
    # my case allows for a children to have many parents. Don't ask.
    children = db.relationship('Child',
                           secondary=parents_children_relationship,
                           backref=db.backref('parents', lazy='dynamic'),
                           lazy='dynamic')

    # many to many relationship between parents and pets
    pets = db.relationship('Pet',
                             secondary=users_pets_relationship,
                             backref=db.backref('parents', lazy='dynamic'), #
                             lazy='dynamic')

# many to many relationship between parents and children
parents_children_relationship = db.Table('parents_children_relationship',
    db.Column('parent_id', db.Integer, db.ForeignKey('parents.id')),
    db.Column('child_id', db.Integer, db.ForeignKey('children.id')),
    UniqueConstraint('parent_id', 'child_id'))

# many to many relationship between User and Pet 
users_pets_relationship = db.Table('users_pets_relationship', 
    db.Column('parent_id', db.Integer, db.ForeignKey('parents.id')), 
    db.Column('pet_id', db.Integer, db.ForeignKey('pets.id')),
    UniqueConstraint('parent_id', 'pet_id'))

class Child(db.Model):
    __tablename__ = 'children'
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(64))
    # parents = <backref relationship with User model>

    # one to many relationship with pets
    pets = db.relationship('Pet', backref='child', lazy='dynamic')


class Pet(db.Model):
    __tablename__ = 'pets'
    id = db.Column(db.Integer, primary_key=True)
    name = db.Column(db.String(64))
    # child = backref relationship with cities
    child_id = db.Column(db.Integer, db.ForeignKey('children.id'), nullable=True)
    # parents = <relationship backref from User>

我想做这样的事情

parent_a = Parent()    
child_a = Child()
pet_a = Pet()

然后我可以这样做

parent_a.children.append(child_a)
# commit/persist data
parent_a.children.all() # [child_a]

我想实现这样的目标

child_a.pets.append(pet_a)
parent_a.children.append(child_a)
# commit/persist data
parent_a.children.all() # [child_a]
parent_a.pets.all() # [pet_a], because pet_a gets 
                    # automatically added to parent using some sorcery
                    # like for child in parent_a.children.all():
                    #     parent.pets.append(child.pets.all())
                    # or something like that.

我可以通过 Parent 对象中的方法实现这一点，例如 add_child_and_its_pets()，但我想覆盖关系的工作方式，所以我不需要覆盖可能受益于此行为的其他模块，例如 Flask-Admin。

基本上我应该如何覆盖 backref.append 方法或 relationship.append 方法以在调用时(即在 python 端)附加来自其他关系的其他对象？我还应该如何覆盖 remove 方法？

Answer 1

对于 parent.pets.all()，我认为您可以将 child 用作 secondary join条件，并将其视为 associative entity or junction table .

这取决于您的表格，但看起来像这样:

Parent.pets = relationship(
    Pet,
    backref='parent'
    primaryjoin=Pet.child_id == Child.id,
    secondaryjoin=Child.parent_id == Parent.id
)

如果您愿意，您也可以相当合理地创建一个 backref parent - 这将使您可以访问 parent_a.pets 和 pet_a.parent.