如何确定哪个函数调用在Python中引发异常？ [英] How to identify what function call raise an exception in Python?

问题描述

我需要确定谁提出了异常来处理更好的str错误，有办法吗？

i need to identify who raise an exception to handle better str error, is there a way ?

看看我的例子：

try:
   os.mkdir('/valid_created_dir')
   os.listdir('/invalid_path')
except OSError, msg:

   # here i want i way to identify who raise the exception
   if is_mkdir_who_raise_an_exception:
      do some things

   if is_listdir_who_raise_an_exception:
      do other things ..

我如何在python中处理呢？

how i can handle this, in python ?

推荐答案

如果您完全依赖于哪个功能执行失败而要执行的任务（如您的代码所示），则尝试使用try /如现有答案所示，exec块可能更好（尽管如果第一个失败，则可能需要跳过第二个部分）。

If you have completely separate tasks to execute depending on which function failed, as your code seems to show, then separate try/exec blocks, as the existing answers suggest, may be better (though you may probably need to skip the second part if the first one has failed).

如果您有很多无论哪种情况，您都需要做一些事情，并且只有很少的工作取决于哪个函数失败，然后分离可能会产生大量的重复和重复，因此您建议的格式可能会更好。 Python标准库中的 traceback 模块可以在此方面提供帮助案例：

If you have many things that you need to do in either case, and only a little amount of work that depends on which function failed, then separating might create a lot of duplication and repetition so the form you suggested may well be preferable. The traceback module in Python's standard library can help in this case:

import os, sys, traceback

try:
   os.mkdir('/valid_created_dir')
   os.listdir('/invalid_path')
except OSError, msg:
   tb = sys.exc_info()[-1]
   stk = traceback.extract_tb(tb, 1)
   fname = stk[0][2]
   print 'The failing function was', fname

当然，如果使用进行检查，您将使用代替打印确切确定要执行的处理。

Of course instead of the print you'll use if checks to decide exactly what processing to do.

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