捕获异常时，如何获取类型、文件和行号?

When I catch an exception, how do I get the type, file, and line number?(捕获异常时，如何获取类型、文件和行号?)

问题描述

捕获会像这样打印的异常:

Catching an exception that would print like this:

Traceback (most recent call last):
  File "c:/tmp.py", line 1, in <module>
    4 / 0
ZeroDivisionError: integer division or modulo by zero

我想把它格式化成:

ZeroDivisonError, tmp.py, 1

推荐答案

import sys, os

try:
    raise NotImplementedError("No error")
except Exception as e:
    exc_type, exc_obj, exc_tb = sys.exc_info()
    fname = os.path.split(exc_tb.tb_frame.f_code.co_filename)[1]
    print(exc_type, fname, exc_tb.tb_lineno)

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