我想为我的应用程序引擎应用程序实现一个简单的VersionedModel基模型类。我正在寻找一种模式,它不需要显式地选择要复制的字段。
我正在尝试这样的东西,但这是为了适应我的口味,还没有在生产环境中进行测试。
class VersionedModel(BaseModel):
is_history_copy = db.BooleanProperty(default=False)
version = db.IntegerProperty()
created = db.DateTimeProperty(auto_now_add=True)
edited = db.DateTimeProperty()
user = db.UserProperty(auto_current_user=True)
def put(self, **kwargs):
if self.is_history_copy:
if self.is_saved():
raise Exception, "History copies of %s are not allowed to change" % type(self).__name__
return super(VersionedModel, self).put(**kwargs)
if self.version is None:
self.version = 1
else:
self.version = self.version +1
self.edited = datetime.now() # auto_now would also affect copies making them out of sync
history_copy = copy.copy(self)
history_copy.is_history_copy = True
history_copy._key = None
history_copy._key_name = None
history_copy._entity = None
history_copy._parent = self
def tx():
result = super(VersionedModel, self).put(**kwargs)
history_copy._parent_key = self.key()
history_copy.put()
return result
return db.run_in_transaction(tx)有没有人有一个更简单的更清洁的解决方案来保存应用程序引擎模型的版本历史?
编辑:将copy移出tx。Thx @Adam Crossland为您提供建议。
发布于 2010-09-13 10:36:38
看看模型类上的属性静态方法。使用此方法,您可以获得属性的列表,并使用该列表获取它们的值,如下所示:
@classmethod
def clone(cls, other, **kwargs):
"""Clones another entity."""
klass = other.__class__
properties = other.properties().items()
kwargs.update((k, p.__get__(other, klass)) for k, p in properties)
return cls(**kwargs)https://stackoverflow.com/questions/3691064
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