SQLAlchemy自引用具有关联代理和对象映射表的多对多

如何解决SQLAlchemy自引用具有关联代理和对象映射表的多对多

我正在尝试在Flask中建立自引用多对多模型。我已经在互联网上检查了很多示例，但就我而言，我想使用对象映射表来处理这种关系。另外，因为我将Flask和棉花糖一起使用，所以我想在表中包括关联代理，以简化模型的序列化。我发现的每个示例都在使用backref，但出于可读性考虑，我想使用back_populates进行创建。目前，我不确定是否可行。 请在下面的最小示例中找到该问题的说明。

from sqlalchemy import Table,Column,Integer,String,ForeignKey
from sqlalchemy.orm import relationship
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.ext.associationproxy import association_proxy
from sqlalchemy import create_engine
from sqlalchemy.orm import sessionmaker

engine = create_engine('sqlite:///:memory:',echo=True)
Session = sessionmaker(bind=engine)
session = Session()

Base = declarative_base()


class NodeRelation(Base):
  __tablename__ = "node_relation"
  parent_id = Column(Integer,ForeignKey('node.id'),primary_key=True)
  parent = relationship('Node',primaryjoin="NodeRelation.parent_id == node.c.id",back_populates='parent_childs',#foreign_keys=parent_id
                        )

  child_id = Column(Integer,primary_key=True)
  child = relationship('Node',primaryjoin="NodeRelation.child_id == node.c.id",back_populates='child_parents',#foreign_keys=child_id
                        )

  def __init__(self,parent=None,child=None,**kwargs):
      super().__init__(**kwargs)
      if child:
          self.child = child
      if parent:
          self.parent = parent

  def __repr__(self):
      return "(parent_id: %s,child_id: %s)" % (self.parent_id,self.child_id)


class Node(Base):
  __tablename__ = "node"
  id = Column(Integer,primary_key=True,autoincrement=True)
  title = Column(String())
  parent_childs = relationship('NodeRelation',primaryjoin="Node.id==node_relation.c.parent_id",back_populates='parent',cascade='all,delete',#foreign_keys=NodeRelation.parent_id
                                )
  parents = association_proxy('parent_childs','parent',creator=lambda parent: NodeRelation(parent=parent))

  child_parents = relationship('NodeRelation',primaryjoin="Node.id==node_relation.c.child_id",back_populates='child',#foreign_keys=NodeRelation.child_id
                                )
  childs = association_proxy('child_parents','child',creator=lambda child: NodeRelation(child=child))

  def __init__(self,title,**kwargs):
    super().__init__(**kwargs)
    self.title = title

  def __repr__(self):
      return "(id: %s,title: %s,childs: %s)" % (self.id,self.title,self.childs)


Base.metadata.create_all(engine)

n1 = Node("First")
n2 = Node("Second")
"""
# This is failing with: NOT NULL constraint failed: node_relation.parent_id
n1.childs.append(n2)

session.add(n1)
session.add(n2)
session.commit()
"""
# This one is working
c = NodeRelation(n1,n2)
session.add(n1)
session.add(n2)
session.add(c)

# Node 1 and Node 2 exists
q = session.query(NodeRelation).all()
print(q)

# This is failing with infinite recursion when childs property is displayed.
q2 = session.query(Node).all()
print(q2)

如果我使用n1.childs.append()，则会出现null约束错误。如果我直接使用n1和n2构造映射器对象，则可以正常工作，但是一旦我访问childs属性，我将获得无限递归。

编辑：

我发现自定义创建者lambda会导致附加错误。因此更新后的代码如下所示：

from sqlalchemy import Table,'parent')

  child_parents = relationship('NodeRelation','child')

  def __init__(self,self.childs)


Base.metadata.create_all(engine)

n1 = Node("First")
n2 = Node("Second")

# This is failing with: NOT NULL constraint failed: node_relation.parent_id
n1.childs.append(n2)

session.add(n1)
session.add(n2)
session.commit()

# Node 1 and Node 2 exists
q = session.query(NodeRelation).all()
print(q)

# This is failing with infinite recursion when childs property is displayed.
q2 = session.query(Node).all()
print(q2)

所以唯一的问题是，当我尝试访问childs属性时，我得到了无限递归。我想我搞砸了一些关系。