Question

如何确定引发的异常函数。例如，存在两个函数：'foo'和'bar'。在'foo'中，异常会随机出现。

import random


def foo():
    if random.randint(1, 10) % 2:
        raise Exception
    bar()


def bar():
    raise Exception

try:
    foo()
except Exception as e:
    print "Exception raised in %s" % ???

Answer 1

import inspect

try:
    foo()
except Exception as e:
    print "Exception raised in %s" % inspect.trace()[-1][3]

Answer 2

我使用跟踪模块，如下所示：

import traceback

try:
    1 / 0
except Exception:
    print traceback.format_exc()

这给出了以下输出：

Traceback (most recent call last):
  File "<ipython-input-3-6b05b5b621cb>", line 2, in <module>
    1 / 0
ZeroDivisionError: integer division or modulo by zero

如果代码从文件运行，则回溯将告诉错误发生的位置的行和字符数：）

修改

以适应Habibutsu：Yes, it's useful for printing, but when needed to get more info (for example function name) - not suitable
的评论
doc-pages告诉您如何以编程方式提取跟踪：http://docs.python.org/2/library/traceback.html

从上面链接的页面：

>>> import traceback >>> def another_function(): ... lumberstack() ... >>> def lumberstack(): ... traceback.print_stack() ... print repr(traceback.extract_stack()) ... print repr(traceback.format_stack()) ... >>> another_function() File "<doctest>", line 10, in <module> another_function() File "<doctest>", line 3, in another_function lumberstack() File "<doctest>", line 6, in lumberstack traceback.print_stack() [('<doctest>', 10, '<module>', 'another_function()'), ('<doctest>', 3, 'another_function', 'lumberstack()'), ('<doctest>', 7, 'lumberstack', 'print repr(traceback.extract_stack())')] [' File "<doctest>", line 10, in <module>\n another_function()\n', ' File "<doctest>", line 3, in another_function\n lumberstack()\n', ' File "<doctest>", line 8, in lumberstack\n print repr(traceback.format_stack())\n']

traceback.extract_stack的文档字符串与traceback.extract_tb
的文档字符串相同

traceback.extract_tb（traceback [，limit]）

返回最多限制“预处理”堆栈跟踪条目的列表   从traceback对象traceback中提取。它很有用   堆栈跟踪的替代格式。如果省略limit或None，则全部   提取条目。 “预处理”堆栈跟踪条目是a   四倍（文件名，行号，函数名，文本）代表   通常为堆栈跟踪打印的信息。文字是   剥离前导和尾随空格的字符串;如果来源   不可用它是无。

Answer 3

你的目标是什么？如果您担心bar和foo抛出相同的异常类型且调用者无法区分它们，只需派生一个新的异常类：

import random

class FooException(Exception):
    """An exception thrown only by foo."""

def foo():
    if random.randint(1,10) % 2:
        raise FooException

    bar()

def bar():
    raise Exception

try:
    foo()
except FooException:
    print "Exception raised in foo..."
except:
    print "Exception raised in bar (probably)..."

如何在Python中找到引发异常的位置

3 个答案: