python - 如何在不停止程序的情况下打印完整的回溯？(How to print the full traceback without halting the program?)

Question

I'm writing a program that parses 10 websites, locates data files, saves the files, and then parses them to make data that can be readily used in the NumPy library. (我正在编写一个程序，该程序可以解析10个网站，找到数据文件，保存文件，然后解析它们以生成可以在NumPy库中轻松使用的数据。) There are tons of errors this file encounters through bad links, poorly formed XML, missing entries, and other things I've yet to categorize. (有过不良链接，不好的XML，缺项，其他的事情我还没有进行分类文件遭遇吨的错误。) I initially made this program to handle errors like this: (我最初制作该程序来处理以下错误：)

try:
    do_stuff()
except:
    pass

But now I want to log errors: (但是现在我想记录错误：)

try:
    do_stuff()
except Exception, err:
    print Exception, err

Note this is printing to a log file for later review. (请注意，这是打印到日志文件以供以后查看。) This usually prints very useless data. (这通常会打印非常无用的数据。) What I want is to print the exact same lines printed when the error triggers without the try-except intercepting the exception, but I don't want it to halt my program since it is nested in a series of for loops that I would like to see to completion. (我想要的是在错误触发时打印完全相同的行，而没有try-except拦截异常，但是我不希望它暂停我的程序，因为它嵌套在我想要的一系列for循环中看到完成。)

  ask by chriscauley translate from so

Answer 1

traceback.format_exc() or sys.exc_info() will yield more info if that's what you want. (如果那是您想要的， traceback.format_exc()或sys.exc_info()将产生更多信息。)

import traceback
import sys

try:
    do_stuff()
except Exception:
    print(traceback.format_exc())
    # or
    print(sys.exc_info()[2])