如何从Python函数中剥离装饰器 [英] How to strip decorators from a function in Python

问题描述

假设我有以下内容：

def with_connection(f):
    def decorated(*args, **kwargs):
        f(get_connection(...), *args, **kwargs)
    return decorated

@with_connection
def spam(connection):
    # Do something

我想测试垃圾邮件函数，而无需经历建立连接的麻烦（或装饰者所做的任何事情）。

I want to test the spam function without going through the hassle of setting up a connection (or whatever the decorator is doing).

给出垃圾邮件，如何从中剥离装饰器并获得基础的未装饰功能？

Given spam, how do I strip the decorator from it and get the underlying "undecorated" function?

推荐答案

一般情况下，您不能这样做，因为

In the general case, you can't, because

@with_connection
def spam(connection):
    # Do something

等同于

def spam(connection):
    # Do something

spam = with_connection(spam)

表示原始 pam可能甚至不存在了。一个（不太漂亮的）黑客是这样的：

which means that the "original" spam might not even exist anymore. A (not too pretty) hack would be this:

def with_connection(f):
    def decorated(*args, **kwargs):
        f(get_connection(...), *args, **kwargs)
    decorated._original = f
    return decorated

@with_connection
def spam(connection):
    # Do something

spam._original(testcon) # calls the undecorated function

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