我在我的代码中有这个try块:

try:
    do_something_that_might_raise_an_exception()
except ValueError as err:
    errmsg = 'My custom error message.'
    raise ValueError(errmsg)

严格地说,我实际上引发了另一个ValueError,而不是do_something…()抛出的ValueError,在这种情况下被称为err。如何将自定义消息附加到错误?我尝试以下代码,但失败,由于错误,ValueError实例,不可调用:

try:
    do_something_that_might_raise_an_exception()
except ValueError as err:
    errmsg = 'My custom error message.'
    raise err(errmsg)

当前回答

It seems all the answers are adding info to e.args[0], thereby altering the existing error message. Is there a downside to extending the args tuple instead? I think the possible upside is, you can leave the original error message alone for cases where parsing that string is needed; and you could add multiple elements to the tuple if your custom error handling produced several messages or error codes, for cases where the traceback would be parsed programmatically (like via a system monitoring tool).

## Approach #1, if the exception may not be derived from Exception and well-behaved:

def to_int(x):
    try:
        return int(x)
    except Exception as e:
        e.args = (e.args if e.args else tuple()) + ('Custom message',)
        raise

>>> to_int('12')
12

>>> to_int('12 monkeys')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in to_int
ValueError: ("invalid literal for int() with base 10: '12 monkeys'", 'Custom message')

or

## Approach #2, if the exception is always derived from Exception and well-behaved:

def to_int(x):
    try:
        return int(x)
    except Exception as e:
        e.args += ('Custom message',)
        raise

>>> to_int('12')
12

>>> to_int('12 monkeys')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 3, in to_int
ValueError: ("invalid literal for int() with base 10: '12 monkeys'", 'Custom message')

你能看出这种方法的缺点吗?

其他回答

try:
    try:
        int('a')
    except ValueError as e:
        raise ValueError('There is a problem: {0}'.format(e))
except ValueError as err:
    print err

打印:

There is a problem: invalid literal for int() with base 10: 'a'

这是我在Python 2.7和3中用来修改异常消息的函数。X,同时保留原始的回溯。需要6个

def reraise_modify(caught_exc, append_msg, prepend=False):
    """Append message to exception while preserving attributes.

    Preserves exception class, and exception traceback.

    Note:
        This function needs to be called inside an except because
        `sys.exc_info()` requires the exception context.

    Args:
        caught_exc(Exception): The caught exception object
        append_msg(str): The message to append to the caught exception
        prepend(bool): If True prepend the message to args instead of appending

    Returns:
        None

    Side Effects:
        Re-raises the exception with the preserved data / trace but
        modified message
    """
    ExceptClass = type(caught_exc)
    # Keep old traceback
    traceback = sys.exc_info()[2]
    if not caught_exc.args:
        # If no args, create our own tuple
        arg_list = [append_msg]
    else:
        # Take the last arg
        # If it is a string
        # append your message.
        # Otherwise append it to the
        # arg list(Not as pretty)
        arg_list = list(caught_exc.args[:-1])
        last_arg = caught_exc.args[-1]
        if isinstance(last_arg, str):
            if prepend:
                arg_list.append(append_msg + last_arg)
            else:
                arg_list.append(last_arg + append_msg)
        else:
            arg_list += [last_arg, append_msg]
    caught_exc.args = tuple(arg_list)
    six.reraise(ExceptClass,
                caught_exc,
                traceback)

这只适用于Python 3。您可以修改异常的原始参数并添加自己的参数。

异常会记住创建它时使用的参数。我认为这样您就可以修改异常了。

在函数rerraise中,我们在异常的原始参数前加上我们想要的任何新参数(比如消息)。最后,我们在保留回溯历史的同时重新引发异常。

def reraise(e, *args):
  '''re-raise an exception with extra arguments
  :param e: The exception to reraise
  :param args: Extra args to add to the exception
  '''

  # e.args is a tuple of arguments that the exception with instantiated with.
  #
  e.args = args + e.args

  # Recreate the exception and preserve the traceback info so that we can see 
  # where this exception originated.
  #
  raise e.with_traceback(e.__traceback__)   


def bad():
  raise ValueError('bad')

def very():
  try:
    bad()
  except Exception as e:
    reraise(e, 'very')

def very_very():
  try:
    very()
  except Exception as e:
    reraise(e, 'very')

very_very()

输出

Traceback (most recent call last):
  File "main.py", line 35, in <module>
    very_very()
  File "main.py", line 30, in very_very
    reraise(e, 'very')
  File "main.py", line 15, in reraise
    raise e.with_traceback(e.__traceback__)
  File "main.py", line 28, in very_very
    very()
  File "main.py", line 24, in very
    reraise(e, 'very')
  File "main.py", line 15, in reraise
    raise e.with_traceback(e.__traceback__)
  File "main.py", line 22, in very
    bad()
  File "main.py", line 18, in bad
    raise ValueError('bad')
ValueError: ('very', 'very', 'bad')

上面提出的许多解决方案再次引发异常,这被认为是一种糟糕的做法。像这样简单的东西就可以了

try:
    import settings
except ModuleNotFoundError:
    print("Something meaningfull\n")
    raise 

因此,您将首先打印错误消息,然后引发堆栈跟踪,或者您可以简单地通过sys.exit(1)退出,而根本不显示错误消息。

当前的答案对我来说并不是很好,如果没有重新捕获异常,则不会显示附加的消息。

但是像下面这样做既保持跟踪,又显示附加的消息,不管是否重新捕获异常。

try:
  raise ValueError("Original message")
except ValueError as err:
  t, v, tb = sys.exc_info()
  raise t, ValueError(err.message + " Appended Info"), tb

(我使用Python 2.7,在Python 3中没有尝试过)