我想在不退出的情况下捕获和记录异常,例如,

try:
    do_stuff()
except Exception as err:
    print(Exception, err)
    # I want to print the entire traceback here,
    # not just the exception name and details

我想打印与抛出异常时打印的完全相同的输出,而不使用try/,只是拦截异常,并且我不希望它退出程序。


当前回答

您需要将try/except放在可能发生错误的最内层循环中,即。

for i in something:
    for j in somethingelse:
        for k in whatever:
            try:
                something_complex(i, j, k)
            except Exception, e:
                print e
        try:
            something_less_complex(i, j)
        except Exception, e:
            print e

... 等等

换句话说,您需要将可能在try/中失败的语句包装在尽可能具体的内部循环中。

其他回答

其他一些答案已经指出了回溯模块。

请注意,使用print_exc,在某些极端情况下,您将无法获得您所期望的结果。在Python 2.x中:

import traceback

try:
    raise TypeError("Oups!")
except Exception, err:
    try:
        raise TypeError("Again !?!")
    except:
        pass

    traceback.print_exc()

...将显示上一个异常的回溯:

Traceback (most recent call last):
  File "e.py", line 7, in <module>
    raise TypeError("Again !?!")
TypeError: Again !?!

如果你真的需要访问原始的回溯,一个解决方案是将exc_info返回的异常信息缓存到一个局部变量中,并使用print_exception显示它:

import traceback
import sys

try:
    raise TypeError("Oups!")
except Exception, err:
    try:
        exc_info = sys.exc_info()

        # do you usefull stuff here
        # (potentially raising an exception)
        try:
            raise TypeError("Again !?!")
        except:
            pass
        # end of useful stuff


    finally:
        # Display the *original* exception
        traceback.print_exception(*exc_info)
        del exc_info

生产:

Traceback (most recent call last):
  File "t.py", line 6, in <module>
    raise TypeError("Oups!")
TypeError: Oups!

但这也有一些陷阱:

From the doc of sys_info: Assigning the traceback return value to a local variable in a function that is handling an exception will cause a circular reference. This will prevent anything referenced by a local variable in the same function or by the traceback from being garbage collected. [...] If you do need the traceback, make sure to delete it after use (best done with a try ... finally statement) but, from the same doc: Beginning with Python 2.2, such cycles are automatically reclaimed when garbage collection is enabled and they become unreachable, but it remains more efficient to avoid creating cycles.


另一方面,通过允许你访问与异常相关的回溯,Python 3产生了一个不那么令人惊讶的结果:

import traceback

try:
    raise TypeError("Oups!")
except Exception as err:
    try:
        raise TypeError("Again !?!")
    except:
        pass

    traceback.print_tb(err.__traceback__)

... 将显示:

  File "e3.py", line 4, in <module>
    raise TypeError("Oups!")

traceback.format_exception (exception_object)

如果你只有异常对象,你可以从Python 3中的任何代码点获得字符串形式的回溯:

import traceback

''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))

完整的例子:

#!/usr/bin/env python3

import traceback

def f():
    g()

def g():
    raise Exception('asdf')

try:
    g()
except Exception as e:
    exc_obj = e

tb_str = ''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))
print(tb_str)

输出:

Traceback (most recent call last):
  File "./main.py", line 12, in <module>
    g()
  File "./main.py", line 9, in g
    raise Exception('asdf')
Exception: asdf

文档:https://docs.python.org/3.9/library/traceback.html traceback.format_exception

请参见:从异常对象中提取回溯信息

在Python 3.9中测试

除了Aaron Hall的回答之外,如果您正在记录日志,但不想使用logging.exception()(因为它在ERROR级别记录日志),您可以使用更低的级别并传递exc_info=True。如。

try:
    do_something_that_might_error()
except Exception:
    logging.info('General exception noted.', exc_info=True)

这是我把错误写在日志文件和控制台的解决方案:

import logging, sys
import traceback
logging.basicConfig(filename='error.log', level=logging.DEBUG)

def handle_exception(exc_type, exc_value, exc_traceback):
    if issubclass(exc_type, KeyboardInterrupt):
        sys.__excepthook__(exc_type, exc_value, exc_traceback)
        return
    exc_info=(exc_type, exc_value, exc_traceback)
    logging.critical("\nDate:" + str(datetime.datetime.now()), exc_info=(exc_type, exc_value, exc_traceback))
    print("An error occured, check error.log to see the error details")
    traceback.print_exception(*exc_info)


sys.excepthook = handle_exception

Python 3解决方案

stacktrace_helper.py:

from linecache import getline
import sys
import traceback


def get_stack_trace():
    exc_type, exc_value, exc_tb = sys.exc_info()
    trace = traceback.format_stack()
    trace = list(filter(lambda x: ("\\lib\\" not in x and "/lib/" not in x and "stacktrace_helper.py" not in x), trace))
    ex_type = exc_type.__name__
    ex_line = exc_tb.tb_lineno
    ex_file = exc_tb.tb_frame.f_code.co_filename
    ex_message = str(exc_value)
    line_code = ""
    try:
        line_code = getline(ex_file, ex_line).strip()
    except:
        pass

    trace.insert(
        0, f'File "{ex_file}", line {ex_line}, line_code: {line_code} , ex: {ex_type} {ex_message}',
    )
    return trace


def get_stack_trace_str(msg: str = ""):
    trace = list(get_stack_trace())
    trace_str = "\n".join(list(map(str, trace)))
    trace_str = msg + "\n" + trace_str
    return trace_str