SqlAlchemy 外键向下传播 3 层

Tim*_*eed 3 python sqlalchemy

我正在尝试更新一个简单的三层关系表集。

他们是

  • 家长
  • 孩子
  • 孙子

该模型的 SQLAlchemy 代码如下所示

class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    children = relationship("Child", back_populates="parents")


class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    parent_id = Column(Integer, ForeignKey('parent.id'))
    parents = relationship("Parent", back_populates="children")
    grandchildren = relationship("GrandChild",
                                back_populates="grandparent",
                                 )

class GrandChild(Base):
    __tablename__ = 'grandchild'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    parent_id = Column(Integer, ForeignKey('parent.id'))
    child_id = Column(Integer, ForeignKey('child.id'))
    grandparent = relationship("Child", back_populates="grandchildren")
Run Code Online (Sandbox Code Playgroud)

插入代码看起来像这样......

p3 = Parent(name="P3")
c5 = Child(name="C5")
c6 = Child(name="C6")
gc1 = GrandChild(name="gc1")
gc2 = GrandChild(name="gc2")
gc3 = GrandChild(name="gc3")
gc4 = GrandChild(name="gc4")

p3.children = [c5, c6]
c5.grandchildren = [gc1]
c6.grandchildren = [gc2,  gc3, gc4]

session.add_all([p2, p3])

session.commit()
Run Code Online (Sandbox Code Playgroud)

记录已添加 - 并且父/子已正确链接 - 但孙子缺少父外键。

我一直在努力寻找正确的机制来添加这个 - 谁能指出我正确的方向?

shm*_*mee 5

您不会在孙子和父母之间建立关系。
孙子和父母之间的关系并不隐含在您的数据模型中;而是隐含在您的数据模型中。孩子的父母不会自动成为孩子所有孙子的父母。

您必须明确定义该关系,即将其添加到GrandChild

class GrandChild(Base):
    [...]
    parent = relationship("Parent")
Run Code Online (Sandbox Code Playgroud)

然后在实例上创建关系:

gc1.parent = p3
gc2.parent = p3
gc3.parent = p3
gc4.parent = p3
Run Code Online (Sandbox Code Playgroud)

这将相应地添加记录:

sqlalchemy.engine.base.Engine INSERT INTO grandchild (name, parent_id, child_id) VALUES (?, ?, ?)
sqlalchemy.engine.base.Engine ('gc1', 1, 1)
[...]
Run Code Online (Sandbox Code Playgroud)

但是,由于数据模型中的父子关系并不意味着任何孙子-父子关系,因此您可以创建一个没有子项但有孙子项的父项。

sink = Parent(name="SINK")
gc1.parent = sink

print("Name: {}, Parent: {}, Parent.children: {}, Child.parent: {}"
      .format(gc1.name, gc1.parent.name, gc1.parent.children, gc1.grandparent.parents.name))
# Name: gc1, Parent: SINK, Parent.children: [], Child.parent: P3
Run Code Online (Sandbox Code Playgroud)

根据我对三层关系的理解,我想不出一个用例。这样就会找到一个应用程序。
如果您希望通过子项在父项和孙项之间建立隐式且一致的关系,请删除父项和孙项之间的直接关系:

class Parent(Base):
    __tablename__ = 'parent'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    children = relationship("Child", back_populates="parent")

    def __repr__(self):
        return "{}(name={})".format(self.__class__.__name__, self.name)

class Child(Base):
    __tablename__ = 'child'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    parent_id = Column(Integer, ForeignKey('parent.id'))
    parent = relationship("Parent", back_populates="children")
    children = relationship("GrandChild", back_populates="parent")

    # same __repr__()

class GrandChild(Base):
    __tablename__ = 'grandchild'
    id = Column(Integer, primary_key=True)
    name = Column(String)
    child_id = Column(Integer, ForeignKey('child.id'))
    parent = relationship("Child", back_populates="children")

    # same __repr__()

p3 = Parent(name="P3")
c5 = Child(name="C5")
gc1 = GrandChild(name="gc1")    
p3.children = [c5]
c5.children = [gc1]
Run Code Online (Sandbox Code Playgroud)

您可以通过以下方式联系孙子的祖父母:

print(gc1.parent.parent)
# Parent(name=P3)
Run Code Online (Sandbox Code Playgroud)

不过,由于层次结构中存在两个一对多关系,另一种方法就有点乏味了:

for child in p3.children:
    for gc in child.children:
        print(p3, child, gc)
# Parent(name=P3) Child(name=C5) GrandChild(name=gc1)
Run Code Online (Sandbox Code Playgroud)