Setting an exit code for a custom exception in python

Question

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

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

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.

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.

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

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

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

Answer 1

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

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