Python:通过具有多个 Excepts 的 Try/Except block 传播异常

Question

有没有办法将 try/except block 中的异常从一个 except 传播到下一个？

我想捕获特定错误，然后再进行一般错误处理。

“引发”是让异常“冒泡”到外部 try/except，但不在引发错误的 try/except block 内。

理想情况下应该是这样的:

import logging

def getList():
    try:
        newList = ["just", "some", "place", "holders"]
        # Maybe from something like: newList = untrustedGetList()

        # Faulty List now throws IndexError
        someitem = newList[100]

        return newList

    except IndexError:
        # For debugging purposes the content of newList should get logged.
        logging.error("IndexError occured with newList containing: \n%s",   str(newList))

    except:
        # General errors should be handled and include the IndexError as well!
        logging.error("A general error occured, substituting newList with backup")
        newList = ["We", "can", "work", "with", "this", "backup"]
        return newList

我遇到的问题是，当 IndexError 被第一个 except 捕获时，我在第二个 except block 中的一般错误处理没有应用。

目前我唯一的解决方法是在第一个 block 中也包含一般错误处理代码。即使我将它包装在它自己的功能 block 中，它仍然看起来不够优雅......

Answer 1

你有两个选择:

• 不要使用专用的 except .. block 捕获 IndexError。您始终可以通过捕获 BaseException 并将异常分配给名称(此处为 e)来手动测试常规 block 中的异常类型:

  try:
      # ...
  except BaseException as e:
      if isinstance(e, IndexError):
          logging.error("IndexError occured with newList containing: \n%s",   str(newList))
  
      logging.error("A general error occured, substituting newList with backup")
      newList = ["We", "can", "work", "with", "this", "backup"]
      return newList
  

• 使用嵌套的 try..except 语句并重新引发:

  try:
      try:
          # ...
      except IndexError:
          logging.error("IndexError occured with newList containing: \n%s",   str(newList))
          raise
  except:
      logging.error("A general error occured, substituting newList with backup")
      newList = ["We", "can", "work", "with", "this", "backup"]
      return newList