默认情况下，将每个函数或类方法包装在Python模块中，而无需修饰每个 [英] Wrap every function or class method in a Python module by default without decorating every one

问题描述

我有一些代码，只要遇到此类异常，我都希望从中获取电子邮件.

I have some code from which I want to get an email whenever it runs into an exception like this.

try:
    f(**kwargs)
except Exception as e:
    # email me the environment

我知道Python装饰器可以做到这一点，例如:

I know that Python decorators can work for this, like:

@check_error
def f()
@check_error
def g()

如果我希望模块中的每个代码默认都被包装怎么办?喜欢没有 @check_error 的 def f()，但是我仍然希望能够实现这一目标.理想情况下，该解决方案应同时适用于函数和类方法.

What if I want every code in my module be wrapped by default? Like def f() without the @check_error, but I still want to be able to achieve this. Ideally, the solution should apply for both functions and class methods.

推荐答案

您可以内省模块，因此假设 module 是要装饰其所有功能的模块，则可以执行以下操作:

You can introspect modules, so assuming module is the module for which you want to decorate all functions you could do something like this:

import inspect
for name, f in inspect.getmembers(module, inspect.isfunction):
    setattr(module, name, check_error(f))

您甚至可以在当前模块中通过对 module

You could even do this in the current module by using sys.modules[__name__] for module

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