假设我以声明方式有三个表,Parent
, Child
, and Pet
,以这样的方式
-
Parent
与两者都有多对多关系Child
and Pet
-
Child
与 具有一对多关系Pet
它们的代码是(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.
我可以通过以下方法实现这一点Parent
类似物体add_child_and_its_pets()
,但我想重写关系的工作方式,所以我不需要重写可能受益于这种行为的其他模块,例如Flask-Admin
例如。
基本上我应该如何覆盖backref.append
方法或relationship.append
方法还可以在调用时附加来自其他关系的其他对象,即在 python 端?我应该如何覆盖remove
方法也一样?