是否可以确定方法是通过属性调用而不是直接调用?
我正在对某些代码进行一些API更改:旧API使用Getters和Setters(GetAttr
和SetAttr
),新的公共API将使用x.Attr
和{{1 }} 分别。我想在程序员调用x.Attr = val
实际上,我正在寻找的是这个神奇的GetAttr()
函数:
_was called_via_property
理想情况下,除了import warnings
class MyClass(object):
def __init__(self):
self._attr = None
def GetAttr(self):
if not _was_called_via_property():
warnings.warn("`GetAttr()` is deprecated. Use `x.attr` property instead.", DeprecationWarning)
return self._attr
def SetAttr(self, value):
if not _was_called_via_property():
warnings.warn("deprecated", DeprecationWarning)
self._attr = value
Attr = property(GetAttr, SetAttr)
函数之外,如果事物是通过装饰器定义的,那么解决方案也会起作用,但这不是必需的。
像这样:
property()
答案 0 :(得分:2)
您无法区分property
描述符访问与直接访问,不能。
创建一个合适的属性,并让旧方法代理它:
@property
def attr(self):
return self._attr
@property.setter
def attr(self, value):
self._attr = value
# legacy access to the attr property
def GetAttr(self):
warnings.warn("deprecated", DeprecationWarning)
return self.attr
def SetAttr(self, value):
warnings.warn("deprecated", DeprecationWarning)
self.attr = value
答案 1 :(得分:1)
另一种解决方案是包裹property
:
def myprop(getter, setter):
return property(lambda self : getter(self, True),
lambda self, x : setter(self, x, True))
class MyClass(object):
def __init__(self):
self._attr = None
def GetAttr(self, called_via_property=False):
if not called_via_property:
warnings.warn("`GetAttr()` is deprecated. Use `x.attr` property instead.", DeprecationWarning)
return self._attr
def SetAttr(self, value, called_via_property=False):
if not called_via_property:
warnings.warn("deprecated", DeprecationWarning)
self._attr = value
Attr = myprop(GetAttr, SetAttr)
另一种解决方案可能是覆盖__getattr__
和__setattr__
以生成带警告的getter和setter,例如:
class MyBase(object):
def __getattr__(self, key):
if key.startswith("Get"):
tail = key[3:]
if hasattr(self, tail):
def getter(self):
res = getattr(self, tail)
issue_warning()
return res
return lambda : getter(self)
raise AttributeError
和setter类似。