当我试图更新使用外键作为主键的一部分的行时遇到了问题。下面是一个简化的情况:
class Foo(Base):
__tablename__ = 'foo_table'
foo_id = Column(Integer, primary_key=True)
bar_id = Column(Integer, ForeignKey('bar_table.bar_id'), primary_key=True)
foo_data = Column(String(255))
bar = relationship('Bar', backref='foos', foreign_keys=[bar_id])
class Bar(Base):
__tablename__ = 'bar_table'
bar_id = Column(Integer, primary_key=True)
首先,我将为foo_table
创建一个条目:
f = Foo()
f.foo_id = 1
f.foo_data = 'Foo Data'
现在,我将在bar_table
中创建一行,并将两者关联起来:
b = Bar()
f.bar = b
很好!我们将把f
添加到会话中并提交:
session.add(f)
session.commit()
现在假设我们遇到了Foo
的另一个示例,它具有相同的foo_id
,并且与相同的Bar
相关,但是具有一些新数据:
f = Foo()
f.foo_id = 1
f.foo_data = 'NEW Foo Data'
f.bar = b
很好!这种情况经常发生,对吧?我将使用session.merge()
而不是session.add()
更新foo_table
中的信息:
session.merge(f)
但这并不好!代码被破解了,我得到了回溯:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/session.py", line 1689, in merge
self._autoflush()
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/session.py", line 1282, in _autoflush
self.flush()
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/session.py", line 2004, in flush
self._flush(objects)
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/session.py", line 2122, in _flush
transaction.rollback(_capture_exception=True)
File "/Library/Python/2.7/site-packages/sqlalchemy/util/langhelpers.py", line 60, in __exit__
compat.reraise(exc_type, exc_value, exc_tb)
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/session.py", line 2086, in _flush
flush_context.execute()
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/unitofwork.py", line 373, in execute
rec.execute(self)
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/unitofwork.py", line 532, in execute
uow
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/persistence.py", line 149, in save_obj
base_mapper, states, uowtransaction
File "/Library/Python/2.7/site-packages/sqlalchemy/orm/persistence.py", line 301, in _organize_states_for_save
state_str(existing)))
sqlalchemy.orm.exc.FlushError: New instance <Foo at 0x10a804590> with identity key (<class 'test.Foo'>, (1, 1)) conflicts with persistent instance <Foo at 0x1097a30d0>
有人知道为什么这个更新失败吗?
2条答案
按热度按时间mi7gmzs61#
我不确定是否有一个真正好的答案......我已经结束了查询,以确定我是否在使用新数据。
所以每当我创建一个新的
Foo
示例时:这看起来并不理想,但我还没有找到另一个有效的解决方案。
l5tcr1uw2#
合并进程中已存在的对象将给予您所看到的错误。
默认情况下,关系启用
cascade='save-update, merge'
设置:save-update
将在调用session.add(object)
时向会话添加任何相关对象。merge
,在与session.merge(object)
合并时将包括任何相关对象。这在SQLAlchemy关于
session.merge()
的文档中有更详细的描述。