我有一个奇怪的问题，我根本无法解决。从本质上讲，我有一个完美的模型和系统——除了在非常具体(似乎是任意的)的情况下。
我马上粘贴模型，但这是我的想法。我希望对某些表进行版本控制。这意味着对于给定的表，我将其分成两个表，Master部分包含对象的自然键，Version表包含所有可能更改的相关数据。然后我的一些模型当然有关系，所以我创建了一个连接表来链接版本。模型如下:
class Versioned(object):
def __init__(self, **kwargs):
super(Versioned, self).__init__(**kwargs)
self.active = True
self.created_on = datetime.datetime.now()
active = Column(BOOLEAN)
created_on = Column(TIMESTAMP, server_default=func.now())
def __eq__(self, other):
return self.__class__ == other.__class__ and
all([getattr(self, key) == getattr(other, key)
for key in self.comparison_keys
])
def __ne__(self, other):
return not self.__eq__(other)
comparison_keys = []
class Parent(Base):
__tablename__ = 'parent'
id = Column(INTEGER, primary_key=True)
name = Column(TEXT)
versions = relationship("ParentVersion", back_populates="master")
children = relationship("Child", back_populates="parent")
@property
def current_version(self):
active_versions = [v for v in self.versions if v.active==True]
return active_versions[0] if active_versions else None
class ParentVersion(Versioned, Base):
__tablename__ = 'parent_version'
id = Column(INTEGER, primary_key=True)
master_id = Column(INTEGER, ForeignKey(Parent.id))
address = Column(TEXT)
master = relationship("Parent", back_populates="versions")
children = relationship("ChildVersion",
secondary=lambda : Parent_Child.__table__
)
class Child(Base):
__tablename__ = 'child'
id = Column(INTEGER, primary_key=True)
parent_id = Column(INTEGER, ForeignKey(Parent.id))
name = Column(TEXT)
versions = relationship("ChildVersion", back_populates="master")
parent = relationship("Parent", back_populates="children")
@property
def current_version(self):
active_versions = [v for v in self.versions if v.active==True]
return active_versions[0] if active_versions else None
class ChildVersion(Versioned, Base):
__tablename__ = 'child_version'
id = Column(INTEGER, primary_key=True)
master_id = Column(INTEGER, ForeignKey(Child.id))
age = Column(INTEGER)
fav_toy = Column(TEXT)
master = relationship("Child", back_populates="versions")
parents = relationship("ParentVersion",
secondary=lambda: Parent_Child.__table__,
)
comparison_keys = [
'age',
'fav_toy',
]
class Parent_Child(Base):
__tablename__ = 'parent_child'
id = Column(INTEGER, primary_key=True)
parent_id = Column(INTEGER, ForeignKey(ParentVersion.id))
child_id = Column(INTEGER, ForeignKey(ChildVersion.id))
好的，我知道最近的SQLAlchemy模型有一些版本控制的想法，我可能用错了方法。但这非常适合我的用例。所以迁就我，让我们假设模型是好的(在一般意义上-如果有一个小细节导致的错误，将是很好的修复)
现在假设我想插入数据。我从某些来源获得数据，然后将其纳入模型。例如，将内容分成Master/Version，分配子关系，分配版本关系。现在我想将它与数据库中的数据进行比较。对于每个主对象，如果我找到它，我会比较它的版本。如果版本不同，则创建一个新版本。棘手的部分是，如果Child版本不同，我想插入一个新的Parent版本，并更新它的所有关系。也许用代码来解释这部分更有意义。search_parent是我在预解析阶段创建的对象。它有一个版本和子对象，子对象也有版本。
parent_conds = [
getattr(search_parent.__class__, name) == getattr(search_parent, name)
for name, column in search_parent.__class__.__mapper__.columns.items()
if not column.primary_key
]
parent_match = session.query(Parent).filter(*parent_conds).first()
# We are going to make a new version
parent_match.current_version.active=False
parent_match.versions.append(search_parent.current_version)
for search_child in search_parent.children[:]:
search_child.parent_id = parent_match.id
search_conds = [
getattr(search_child.__class__, name) == getattr(search_child, name)
for name, column in search_child.__class__.__mapper__.columns.items()
if not column.primary_key
]
child_match = session.query(Child).filter(*search_conds).first()
if child_match.current_version != search_child.current_version:
# create a new version: deactivate the old one, insert the new
child_match.current_version.active=False
child_match.versions.append(search_child.current_version)
else:
# copy the old version to point to the new parent version
children = parent_match.current_version.children
children.append(child_match.current_version)
children.remove(search_child.current_version)
session.expunge(search_child.current_version)
session.expunge(search_child)
session.expunge(search_parent)
session.add(parent_match)
session.commit()
好了，再一次，这可能不是完美的，甚至不是最好的方法。但它确实有效。除了，这就是我不明白的。如果我将孩子的age属性更新为整数值0，它将不起作用。如果子对象从0岁开始，我把它改成别的，效果会很好。如果我从某个非零整数开始，并将年龄更新为0，我会得到以下警告:
SAWarning: Object of type <ChildVersion> not in session, add operation along 'ParentVersion.children' won't proceed (mapperutil.state_class_str(child), operation, self.prop))
插入更新后的版本，但是没有插入到parent_child连接表中。这并不是说它失败了，而是SQLAlchemy确定子对象不存在，因此无法创建连接。但它确实存在，我知道它被插入了
同样，只有在插入age=0的新版本时才会发生这种情况。如果我插入一个其他版本的新版本，它就会像我想要的那样工作。
关于这个bug还有其他奇怪的事情——如果你没有插入足够的子元素(似乎大约12个会触发这个bug)，它不会发生，有时取决于其他属性。我不认为我完全理解是什么引起了它的表面区域。
感谢您花时间阅读这篇文章。我有一个完整的工作演示和源数据，我很乐意分享，它只是需要一些设置，所以我不知道它是否适合这篇文章。我希望有人能告诉我该看什么，因为在这一点上我完全不知道。
edit:这是导致警告的完整堆栈跟踪。
File "repro.py", line 313, in <module>
load_data(session, second_run)
File "repro.py", line 293, in load_data
session.commit()
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/session.py", line 801, in commit
self.transaction.commit()
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/session.py", line 392, in commit
self._prepare_impl()
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/session.py", line 372, in _prepare_impl
self.session.flush()
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/session.py", line 2019, in flush
self._flush(objects)
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/session.py", line 2101, in _flush
flush_context.execute()
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/unitofwork.py", line 373, in execute
rec.execute(self)
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/unitofwork.py", line 487, in execute
self.dependency_processor.process_saves(uow, states)
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/dependency.py", line 1053, in process_saves
False, uowcommit, "add"):
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/dependency.py", line 1154, in _synchronize
(mapperutil.state_class_str(child), operation, self.prop))
File "/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/util/langhelpers.py", line 1297, in warn
warnings.warn(msg, exc.SAWarning, stacklevel=2)
File "repro.py", line 10, in warn_with_traceback
traceback.print_stack()
/Users/me/virtualenvs/dev/lib/python2.7/site-packages/sqlalchemy/orm/dependency.py:1154: SAWarning: Object of type <ChildVersion> not in session, add operation along 'ParentVersion.children' won't proceed
(mapperutil.state_class_str(child), operation, self.prop))
edit2:下面是一个python文件的要点，您可以运行以查看奇怪的行为。https://gist.github.com/jbouricius/2ede420fb1f7a2deec9f557c76ced7f9
得到此错误的原因是您无意中向会话中添加了对象。
MVCE:
engine = create_engine("sqlite://", echo=False)
def get_data():
children = [
Child(name="Carol", versions=[ChildVersion(age=0, fav_toy="med")]),
Child(name="Timmy", versions=[ChildVersion(age=0, fav_toy="med")]),
]
return Parent(
name="Zane", children=children,
versions=[
ParentVersion(
address="123 Fake St",
children=[v for child in children for v in child.versions]
)
]
)
def main():
Base.metadata.create_all(engine)
session = Session(engine)
parent_match = get_data()
session.add(parent_match)
session.commit()
with session.no_autoflush:
search_parent = get_data()
parent_match.versions.append(search_parent.current_version)
for search_child in search_parent.children[:]:
child_match = next(c for c in parent_match.children if c.name == search_child.name)
if child_match.current_version != search_child.current_version:
child_match.versions.append(search_child.current_version)
else:
session.expunge(search_child.current_version)
session.expunge(search_child)
session.expunge(search_parent)
session.commit()
旁白:这是你需要在问题本身中提供的内容。提供带有指令的tarball文件并不是获得答案的最佳方式。
行parent_match.versions.append(search_parent.current_version)
不仅添加了search_parent.current_version，还添加了search_parent，这反过来又添加了所有相关对象，包括其他子对象的子版本。根据您稍后删除其他相关对象以防止它们被添加到会话的事实判断，我得出的结论是您只想添加search_parent.current_version而不添加其他相关对象。由于关系的循环性质，在添加对象之前，您需要注意只从search_parent中取出您想要的对象。下面是固定的MVCE:
with session.no_autoflush:
search_parent = get_data()
current_parent_version = search_parent.current_version
search_parent.versions.remove(current_parent_version)
current_parent_version.children = [] # <--- this is key
for search_child in search_parent.children[:]:
child_match = next(c for c in parent_match.children if c.name == search_child.name)
if child_match.current_version != search_child.current_version:
current_child_version = search_child.current_version
search_child.versions.remove(current_child_version)
child_match.versions.append(current_child_version)
current_parent_version.children.append(current_child_version)
parent_match.versions.append(current_parent_version)
session.commit()