我想从数据库中获取一个已经存在的对象(基于提供的参数)，或者创建一个不存在的对象.
Django的get_or_create(或source)可以做到这一点.炼金术中有没有类似的捷径？
我现在明确地这样写:
def get_or_create_instrument(session, serial_number):
instrument = session.query(Instrument).filter_by(serial_number=serial_number).first()
if instrument:
return instrument
else:
instrument = Instrument(serial_number)
session.add(instrument)
return instrument
基本上就是这么做的，没有现成的捷径.
当然，你可以概括如下:
def get_or_create(session, model, defaults=None, **kwargs):
instance = session.query(model).filter_by(**kwargs).one_or_none()
if instance:
return instance, False
else:
params = {k: v for k, v in kwargs.items() if not isinstance(v, ClauseElement)}
params.update(defaults or {})
instance = model(**params)
try:
session.add(instance)
session.commit()
except Exception: # The actual exception depends on the specific database so we catch all exceptions. This is similar to the official documentation: https://docs.sqlalchemy.org/en/latest/orm/session_transaction.html
session.rollback()
instance = session.query(model).filter_by(**kwargs).one()
return instance, False
else:
return instance, True
这是一个使用Python 3.9的the new dict union operator (|=)的更简洁版本
def get_or_create(session, model, defaults=None, **kwargs):
instance = session.query(model).filter_by(**kwargs).one_or_none()
if instance:
return instance, False
else:
kwargs |= defaults or {}
instance = model(**kwargs)
try:
session.add(instance)
session.commit()
except Exception: # The actual exception depends on the specific database so we catch all exceptions. This is similar to the official documentation: https://docs.sqlalchemy.org/en/latest/orm/session_transaction.html
session.rollback()
instance = session.query(model).filter_by(**kwargs).one()
return instance, False
else:
return instance, True
与Django版本类似，这将捕获重复的键约束和类似的错误.如果您的GET或CREATE不能保证返回单个结果，它仍然可能导致争用条件.
要缓解某些问题，您需要在session.commit()后面添加另一个one_or_none()样式的FETCH.除非您还使用with_for_update()或可序列化事务模式，否则对于争用条件，这仍然不是one_or_none()%的保证.