在python中为自定义异常设置退出代码 [英] Setting an exit code for a custom exception in python

问题描述

我正在使用自定义异常将我的异常与Python的默认异常区分开。

I'm using custom exceptions to differ my exceptions from Python's default exceptions.

当引发异常时，是否可以定义自定义退出代码？

Is there a way to define a custom exit code when I raise the exception?

class MyException(Exception):
    pass

def do_something_bad():
    raise MyException('This is a custom exception')

if __name__ == '__main__':
    try:
        do_something_bad()
    except:
        print('Oops')  # Do some exception handling
        raise

在此代码中， main函数在try代码中运行一些函数。
捕获异常后，我想重新引发它以保留回溯堆栈。

In this code, the main function runs a few functions in a try code. After I catch an exception I want to re-raise it to preserve the traceback stack.

问题是'raise'总是退出1。
我想使用自定义退出代码（针对我的自定义异常）退出脚本，并在其他情况下退出1。

The problem is that 'raise' always exits 1. I want to exit the script with a custom exit code (for my custom exception), and exit 1 in any other case.

我已经看过此解决方案但这不是我想要的东西：
在引发异常时在Python中设置退出代码

I've looked at this solution but it's not what I'm looking for: Setting exit code in Python when an exception is raised

此解决方案迫使我检查使用的每个脚本中的默认值还是默认值

This solution forces me to check in every script I use whether the exception is a default or a custom one.

我希望我的自定义异常能够告诉抬高函数使用哪种退出代码。

I want my custom exception to be able to tell the raise function what exit code to use.

推荐答案

您可以覆盖 sys.excepthook 做自己想做的事情：

You can override sys.excepthook to do what you want yourself:

import sys

class ExitCodeException(Exception):
  "base class for all exceptions which shall set the exit code"
  def getExitCode(self):
    "meant to be overridden in subclass"
    return 3

def handleUncaughtException(exctype, value, trace):
  oldHook(exctype, value, trace)
  if isinstance(value, ExitCodeException):
    sys.exit(value.getExitCode())

sys.excepthook, oldHook = handleUncaughtException, sys.excepthook

这样，您可以将此代码放在一个特殊的模块中您所有的代码都只需要导入。

This way you can put this code in a special module which all your code just needs to import.

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