将关联对象与 back_populates 一起使用时，SQLAlchemy 抛出 KeyError – 文档中的示例不起作用

SQLAlchemy 很好的文档如何使用关联对象back_populates.

但是，当从该文档复制并粘贴示例时，将子项添加到父项会引发KeyError如下面的代码所示。模型类 100% 从文档中复制：

from sqlalchemy import Column, ForeignKey, Integer, String
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import relationship
from sqlalchemy.schema import MetaData

Base = declarative_base(metadata=MetaData())

class Association(Base):
    __tablename__ = 'association'
    left_id = Column(Integer, ForeignKey('left.id'), primary_key=True)
    right_id = Column(Integer, ForeignKey('right.id'), primary_key=True)
    extra_data = Column(String(50))
    child = relationship("Child", back_populates="parents")
    parent = relationship("Parent", back_populates="children")

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

class Child(Base):
    __tablename__ = 'right'
    id = Column(Integer, primary_key=True)
    parents = relationship("Association", back_populates="child")

parent = Parent(children=[Child()])

使用 SQLAlchemy 版本 1.2.11 运行该代码会引发以下异常：

lars$ venv/bin/python test.py
Traceback (most recent call last):
  File "test.py", line 26, in <module>
    parent = Parent(children=[Child()])
  File "<string>", line 4, in __init__
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/state.py", line 417, in _initialize_instance
    manager.dispatch.init_failure(self, args, kwargs)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/util/langhelpers.py", line 66, in __exit__
    compat.reraise(exc_type, exc_value, exc_tb)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/util/compat.py", line 249, in reraise
    raise value
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/state.py", line 414, in _initialize_instance
    return manager.original_init(*mixed[1:], **kwargs)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/ext/declarative/base.py", line 737, in _declarative_constructor
    setattr(self, k, kwargs[k])
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/attributes.py", line 229, in __set__
    instance_dict(instance), value, None)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/attributes.py", line 1077, in set
    initiator=evt)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/collections.py", line 762, in bulk_replace
    appender(member, _sa_initiator=initiator)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/collections.py", line 1044, in append
    item = __set(self, item, _sa_initiator)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/collections.py", line 1016, in __set
    item = executor.fire_append_event(item, _sa_initiator)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/collections.py", line 680, in fire_append_event
    item, initiator)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/attributes.py", line 943, in fire_append_event
    state, value, initiator or self._append_token)
  File "/Users/lars/coding/sqlalchemy_association_object_test/venv/lib/python3.7/site-packages/sqlalchemy/orm/attributes.py", line 1210, in emit_backref_from_collection_append_event
    child_impl = child_state.manager[key].impl
KeyError: 'parent'

我已将其归档为SQLAlchemy 问题跟踪器中的错误。也许有人可以同时向我指出一个可行的解决方案或解决方法？

tldr;我们必须使用关联代理扩展and为关联对象创建一个自定义构造函数，它将子对象作为第一个 (!) 参数。请参阅基于以下问题示例的解决方案。

SQLAlchemy 的文档实际上在下一段中指出，如果我们想直接添加，我们还没有完成Child模型到Parent跳过中介的模型Association models:

以直接形式使用关联模式需要： 子对象在被关联之前先与关联实例关联 附加到父级；同样，从父母到孩子的访问 通过关联对象。

# create parent, append a child via association
p = Parent()
a = Association(extra_data="some data")
a.child = Child()
p.children.append(a)

编写问题中所要求的方便的代码，即p.children = [Child()]，我们必须利用关联代理扩展.

这是使用关联代理扩展的解决方案，它允许“直接”将子级添加到父级，而无需在它们之间显式创建关联：

from sqlalchemy import Column, ForeignKey, Integer, String
from sqlalchemy.ext.associationproxy import association_proxy
from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy.orm import backref, relationship
from sqlalchemy.schema import MetaData

Base = declarative_base(metadata=MetaData())

class Association(Base):
    __tablename__ = 'association'
    left_id = Column(Integer, ForeignKey('left.id'), primary_key=True)
    right_id = Column(Integer, ForeignKey('right.id'), primary_key=True)
    extra_data = Column(String(50))
    child = relationship("Child", back_populates="parents")
    parent = relationship("Parent", backref=backref("parent_children"))

    def __init__(self, child=None, parent=None):
        self.parent = parent
        self.child = child

class Parent(Base):
    __tablename__ = 'left'
    id = Column(Integer, primary_key=True)
    children = association_proxy("parent_children", "child")

class Child(Base):
    __tablename__ = 'right'
    id = Column(Integer, primary_key=True)
    parents = relationship("Association", back_populates="child")

p = Parent(children=[Child()])

不幸的是我只知道如何使用backref代替back_populates这不是“现代”方法。

特别注意创建自定义__init__以孩子为对象的方法first争论。