为什么 hasattr 会执行@property 装饰器代码块 [英] Why does hasattr execute the @property decorator code block

问题描述

在 Python 中，当我在 @property 装饰器上调用 hasattr 时，hasattr 函数实际上运行了 @property 代码块.

In Python when I call the hasattr on a @property decorator the hasattr function actually runs the @property code block.

例如一个类:

class GooglePlusUser(object):

    def __init__(self, master):
        self.master = master

    def get_user_id(self):
        return self.master.google_plus_service.people().get(userId='me').execute()['id']


    @property
    def profile(self):
        # this runs with hasattr
        return self.master.google_plus_service.people().get(userId='me').execute()

运行以下代码调用 profile 属性并实际进行调用:

Running the following code calls the profile property and actually makes the call:

#Check if the call is an attribute
if not hasattr(google_plus_user, call):
    self.response.out.write('Unknown call')
    return

为什么?如何在不进行 api 调用的情况下解决此问题?

Why? How can I solve this without making the api call?

推荐答案

hasattr() 通过实际检索属性来工作；如果抛出异常 hasattr() 返回 False.这是因为这是知道属性是否存在的唯一可靠方法，因为有很多动态方法可以在 Python 对象上注入属性(__getattr__、__getattribute__、属性 对象、元类等).

hasattr() works by actually retrieving the attribute; if an exception is thrown hasattr() returns False. That's because that is the only reliable way of knowing if an attribute exists, since there are so many dynamic ways to inject attributes on Python objects (__getattr__, __getattribute__, property objects, meta classes, etc.).

来自 hasattr() 文档:

From the hasattr() documentation:

这是通过调用getattr(object, name) 并查看它是否引发异常来实现的.

This is implemented by calling getattr(object, name) and seeing whether it raises an exception or not.

如果您不希望在执行此操作时调用某个属性，则不要使用 hasattr.使用 vars()(返回实例字典)或 dir()(它还为您提供类的名称列表).

If you don't want a property to be invoked when doing this, then don't use hasattr. Use vars() (which returns the instance dictionary) or dir() (which gives you a list of names on the class as well).

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