python - 如何在Python中缩进堆栈跟踪？

Question

环境 - win7 机器上的 Python 2.7.2。我的技能水平 - 菜鸟

我正在使用以下内容来捕获并打印异常堆栈跟踪 -

def logerr(stmt, e):
try:
    ##do something

except:
    print '##EXCEPTION in logging: '
    exc_type, exc_value, exc_traceback = sys.exc_info()
    traceback.print_exception(exc_type, exc_value, exc_traceback, file=sys.stdout)

输出是-

##EXCEPTION in logging: 
Traceback (most recent call last):
  File "C:\Users\amurty\Desktop\dev\eclipse\workspace\hhs\FeedSearch\src\main\main.py", line 18, in main
    log()
TypeError: log() takes exactly 1 argument (0 given)

我想缩进堆栈跟踪。所以输出应该是这样的 -

##EXCEPTION in logging: 
        Traceback (most recent call last):
          File "C:\Users\amurty\Desktop\dev\eclipse\workspace\hhs\FeedSearch\src\main\main.py", line 18, in main
            log()
        TypeError: log() takes exactly 1 argument (0 given)

我怎样才能实现这个目标。pprint 或 textwrap 模块在这里有帮助吗？

Answer 1

试试这个:

import traceback

def logerr(stmt, e):
    try:
        ##do something

    except:
        print '##EXCEPTION in logging: '
        for line in traceback.format_exception().splitlines():
            print '        ' + line