如何在不停止程序的情况下打印完整的回溯?

当你不想因为一个错误而停止你的程序时，你需要用try/except来处理这个错误:

try:
    do_something_that_might_error()
except Exception as error:
    handle_the_error(error)

要提取完整的回溯，我们将使用标准库中的traceback模块:

import traceback

并创建一个相当复杂的stacktrace来演示我们得到完整的stacktrace:

def raise_error():
    raise RuntimeError('something bad happened!')

def do_something_that_might_error():
    raise_error()

印刷

要打印完整的回溯，请使用traceback。print_exc方法:

try:
    do_something_that_might_error()
except Exception as error:
    traceback.print_exc()

打印:

Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
  File "<stdin>", line 2, in do_something_that_might_error
  File "<stdin>", line 2, in raise_error
RuntimeError: something bad happened!

优于打印、日志:

但是，最佳实践是为您的模块设置一个记录器。它将知道模块的名称，并且能够更改级别(在其他属性中，例如处理程序)

import logging
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)

在这种情况下，您将需要记录器。异常函数改为:

try:
    do_something_that_might_error()
except Exception as error:
    logger.exception(error)

日志:

ERROR:__main__:something bad happened!
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
  File "<stdin>", line 2, in do_something_that_might_error
  File "<stdin>", line 2, in raise_error
RuntimeError: something bad happened!

或者，您可能只想要字符串，在这种情况下，您将需要回溯。Format_exc函数改为:

try:
    do_something_that_might_error()
except Exception as error:
    logger.debug(traceback.format_exc())

日志:

DEBUG:__main__:Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
  File "<stdin>", line 2, in do_something_that_might_error
  File "<stdin>", line 2, in raise_error
RuntimeError: something bad happened!

结论

对于这三个选项，我们看到我们得到的输出和我们有错误时是一样的:

>>> do_something_that_might_error()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in do_something_that_might_error
  File "<stdin>", line 2, in raise_error
RuntimeError: something bad happened!

使用哪种

性能问题在这里并不重要，因为IO通常占主导地位。我更喜欢，因为它精确地以向前兼容的方式执行请求:

logger.exception(error)

日志级别和输出可以调整，这样就可以很容易地关闭，而不需要修改代码。通常做直接需要的是最有效的方法。

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)

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

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。

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--------

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!")