Question

SQLAlchemy很好地记录了how to use Association Objects with back_populates。

但是，当从该文档中复制并粘贴示例时，将子级添加到父级会抛出KeyError，如以下代码所示。从文档中100％复制模型类：

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 1.2.11版运行该代码会引发此异常：

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bug in SQLAlchemy's issue tracker。也许有人可以同时向我指出可行的解决方案或解决方法？

Answer 1

tldr; 我们必须使用关联代理扩展，并且为关联对象创建一个自定义构造函数，该构造函数将子对象作为第一个（！）参数。请根据以下问题的示例查看解决方案。

SQLAlchemy的文档实际上在下一段中指出，如果我们想直接将Child模型添加到Parent模型中，而跳过中间Association模型，则我们还没有完成：

使用直接形式的关联模式需要子对象先与关联实例关联附于父母之后；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)

要编写问题中要求的便捷代码，即p.children = [Child()]，我们必须使用Association Proxy extension。

这是使用关联代理扩展的解决方案，该扩展允许“直接”将子代添加到父代，而无需显式地创建两个子代之间的关联：

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，我只是想出了如何使用backref而不是back_populates的方法，这不是“现代”方法。

要特别注意创建一个自定义__init__方法，该方法将孩子作为 first 参数。

Answer 2

所以总而言之。

您需要将包含子对象的关联对象附加到父对象上。否则，您需要遵循Lars关于代理关联的建议。

我推荐前者，因为它是基于ORM的方式：

p = Parent()
p.children.append(Association(child = Child()))
session.add(p)
session.commit()

请注意，如果您有任何非空字段，则可以在创建对象时轻松添加它们以进行快速测试提交。

结合使用带有Back_populates的关联对象时，SQLAlchemy抛出KeyError –文档中的示例不起作用

2 个答案: