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

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

其他回答

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)

在python3(适用于3.9)中,我们可以定义一个函数,并可以在任何需要打印详细信息的地方使用它。

import traceback

def get_traceback(e):
    lines = traceback.format_exception(type(e), e, e.__traceback__)
    return ''.join(lines)

try:
    1/0
except Exception as e:
    print('------Start--------')
    print(get_traceback(e))
    print('------End--------')

try:
    spam(1,2)
except Exception as e:
    print('------Start--------')
    print(get_traceback(e))
    print('------End--------')

输出如下所示:

bash-3.2$ python3 /Users/soumyabratakole/PycharmProjects/pythonProject/main.py
------Start--------
Traceback (most recent call last):
  File "/Users/soumyabratakole/PycharmProjects/pythonProject/main.py", line 26, in <module>
    1/0
ZeroDivisionError: division by zero

------End--------
------Start--------
Traceback (most recent call last):
  File "/Users/soumyabratakole/PycharmProjects/pythonProject/main.py", line 33, in <module>
    spam(1,2)
NameError: name 'spam' is not defined

------End--------

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。

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

我在其他答案中没有看到这个。如果你出于某种原因传递一个Exception对象……

在Python 3.5+中,您可以使用traceback.TracebackException.from_exception()从Exception对象获取跟踪。例如:

import traceback


def stack_lvl_3():
    raise Exception('a1', 'b2', 'c3')


def stack_lvl_2():
    try:
        stack_lvl_3()
    except Exception as e:
        # raise
        return e


def stack_lvl_1():
    e = stack_lvl_2()
    return e

e = stack_lvl_1()

tb1 = traceback.TracebackException.from_exception(e)
print(''.join(tb1.format()))

然而,上面的代码导致:

Traceback (most recent call last):
  File "exc.py", line 10, in stack_lvl_2
    stack_lvl_3()
  File "exc.py", line 5, in stack_lvl_3
    raise Exception('a1', 'b2', 'c3')
Exception: ('a1', 'b2', 'c3')

这只是堆栈的两层,而不是在stack_lvl_2()中引发异常而没有被拦截(取消注释# raise行)时在屏幕上打印的内容。

根据我的理解,这是因为异常在被引发时只记录堆栈的当前级别,在本例中是stack_lvl_3()。当它在堆栈中往回传递时,更多的层被添加到它的__traceback__中。但是我们在stack_lvl_2()中拦截了它,这意味着它只能记录级别3和2。要获得打印在stdout上的完整跟踪,我们必须在最高(最低?)级别捕获它:

import traceback


def stack_lvl_3():
    raise Exception('a1', 'b2', 'c3')


def stack_lvl_2():
    stack_lvl_3()


def stack_lvl_1():
    stack_lvl_2()


try:
    stack_lvl_1()
except Exception as exc:
    tb = traceback.TracebackException.from_exception(exc)

print('Handled at stack lvl 0')
print(''.join(tb.stack.format()))

结果是:

Handled at stack lvl 0
  File "exc.py", line 17, in <module>
    stack_lvl_1()
  File "exc.py", line 13, in stack_lvl_1
    stack_lvl_2()
  File "exc.py", line 9, in stack_lvl_2
    stack_lvl_3()
  File "exc.py", line 5, in stack_lvl_3
    raise Exception('a1', 'b2', 'c3')

注意,堆栈打印是不同的,第一行和最后一行都不见了。因为它是不同的格式()。

在尽可能远离异常引发点的地方拦截异常,可以简化代码,同时提供更多信息。