捕获输出如下的异常:

Traceback (most recent call last):
  File "c:/tmp.py", line 1, in <module>
    4 / 0
ZeroDivisionError: integer division or modulo by zero

我想把它格式化为:

ZeroDivisonError, tmp.py, 1

当前回答

traceback.format_exception()的源代码(Py v2.7.3)和调用的/相关函数有很大帮助。令人尴尬的是,我总是忘记阅读原文。我只是在徒劳地搜索了类似的细节后才这样做的。一个简单的问题,“如何为异常重新创建与Python相同的输出,并具有所有相同的细节?”这将使任何人得到90%以上的他们想要的东西。沮丧的我想出了这个例子。我希望它能帮助其他人。(它确实帮助了我!: -)

import sys, traceback

traceback_template = '''Traceback (most recent call last):
  File "%(filename)s", line %(lineno)s, in %(name)s
%(type)s: %(message)s\n''' # Skipping the "actual line" item

# Also note: we don't walk all the way through the frame stack in this example
# see hg.python.org/cpython/file/8dffb76faacc/Lib/traceback.py#l280
# (Imagine if the 1/0, below, were replaced by a call to test() which did 1/0.)

try:
    1/0
except:
    # http://docs.python.org/2/library/sys.html#sys.exc_info
    exc_type, exc_value, exc_traceback = sys.exc_info() # most recent (if any) by default

    '''
    Reason this _can_ be bad: If an (unhandled) exception happens AFTER this,
    or if we do not delete the labels on (not much) older versions of Py, the
    reference we created can linger.

    traceback.format_exc/print_exc do this very thing, BUT note this creates a
    temp scope within the function.
    '''

    traceback_details = {
                         'filename': exc_traceback.tb_frame.f_code.co_filename,
                         'lineno'  : exc_traceback.tb_lineno,
                         'name'    : exc_traceback.tb_frame.f_code.co_name,
                         'type'    : exc_type.__name__,
                         'message' : exc_value.message, # or see traceback._some_str()
                        }

    del(exc_type, exc_value, exc_traceback) # So we don't leave our local labels/objects dangling
    # This still isn't "completely safe", though!
    # "Best (recommended) practice: replace all exc_type, exc_value, exc_traceback
    # with sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]

    print
    print traceback.format_exc()
    print
    print traceback_template % traceback_details
    print

具体回答此问题:

sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno

其他回答

traceback.format_exception()的源代码(Py v2.7.3)和调用的/相关函数有很大帮助。令人尴尬的是,我总是忘记阅读原文。我只是在徒劳地搜索了类似的细节后才这样做的。一个简单的问题,“如何为异常重新创建与Python相同的输出,并具有所有相同的细节?”这将使任何人得到90%以上的他们想要的东西。沮丧的我想出了这个例子。我希望它能帮助其他人。(它确实帮助了我!: -)

import sys, traceback

traceback_template = '''Traceback (most recent call last):
  File "%(filename)s", line %(lineno)s, in %(name)s
%(type)s: %(message)s\n''' # Skipping the "actual line" item

# Also note: we don't walk all the way through the frame stack in this example
# see hg.python.org/cpython/file/8dffb76faacc/Lib/traceback.py#l280
# (Imagine if the 1/0, below, were replaced by a call to test() which did 1/0.)

try:
    1/0
except:
    # http://docs.python.org/2/library/sys.html#sys.exc_info
    exc_type, exc_value, exc_traceback = sys.exc_info() # most recent (if any) by default

    '''
    Reason this _can_ be bad: If an (unhandled) exception happens AFTER this,
    or if we do not delete the labels on (not much) older versions of Py, the
    reference we created can linger.

    traceback.format_exc/print_exc do this very thing, BUT note this creates a
    temp scope within the function.
    '''

    traceback_details = {
                         'filename': exc_traceback.tb_frame.f_code.co_filename,
                         'lineno'  : exc_traceback.tb_lineno,
                         'name'    : exc_traceback.tb_frame.f_code.co_name,
                         'type'    : exc_type.__name__,
                         'message' : exc_value.message, # or see traceback._some_str()
                        }

    del(exc_type, exc_value, exc_traceback) # So we don't leave our local labels/objects dangling
    # This still isn't "completely safe", though!
    # "Best (recommended) practice: replace all exc_type, exc_value, exc_traceback
    # with sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]

    print
    print traceback.format_exc()
    print
    print traceback_template % traceback_details
    print

具体回答此问题:

sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno

这是我用来获取文件名的。

__file__.__str__

总之,我创建了一个页面来显示错误。返回此异常。

context={
                'details':'Type of error:{}:Function name:{}:Line number:{}'.format(exc_type, fname, exc_tb.tb_lineno),
                'error_details':str(e),
                'filename':__file__.__str__,
            })

发生异常时的情况:

import sys, os

try:
    raise NotImplementedError("No error")
except Exception as e:
    exc_type, exc_obj, exc_tb = sys.exc_info()
    fname = os.path.split(exc_tb.tb_frame.f_code.co_filename)[1]
    print(exc_type, fname, exc_tb.tb_lineno)

没有任何导入,但也不兼容导入的模块:

try:
    raise TypeError("Hello, World!")  # line 2
except Exception as e:
    print(
        type(e).__name__,          # TypeError
        __file__,                  # /tmp/example.py
        e.__traceback__.tb_lineno  # 2
    )

$ python3 /tmp/example.py TypeError /tmp/example.py

重申一下,这不能跨导入或模块工作,所以如果你导入X;试题:X.example ();那么文件名和行号将指向包含X.example()的行,而不是X.example()中出错的行。如果有人知道如何轻松地从最后一个堆栈跟踪行中获得文件名和行号(我期望类似e[-1]的内容)。文件名,但没有这样的运气),请改进这个答案。

下面是一个显示发生异常的行号的示例。

import sys
try:
    print(5/0)
except Exception as e:
    print('Error on line {}'.format(sys.exc_info()[-1].tb_lineno), type(e).__name__, e)

print('And the rest of program continues')