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社区首页 >问答首页 >当我捕获到异常时,如何获取类型、文件和行号?

当我捕获到异常时,如何获取类型、文件和行号?
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Stack Overflow用户
提问于 2009-08-14 16:02:15
回答 6查看 315.4K关注 0票数 335

捕获如下所示的异常输出:

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

我想将其格式化为:

代码语言:javascript
复制
ZeroDivisonError, tmp.py, 1
EN

回答 6

Stack Overflow用户

回答已采纳

发布于 2009-08-14 16:09:06

代码语言:javascript
复制
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)
票数 453
EN

Stack Overflow用户

发布于 2017-12-06 01:10:30

对我有效的最简单的形式。

代码语言:javascript
复制
import traceback

try:
    print(4/0)
except ZeroDivisionError:
    print(traceback.format_exc())

输出

代码语言:javascript
复制
Traceback (most recent call last):
  File "/path/to/file.py", line 51, in <module>
    print(4/0)
ZeroDivisionError: division by zero

Process finished with exit code 0
票数 232
EN

Stack Overflow用户

发布于 2013-04-17 04:55:32

用于traceback.format_exception()和被调用/相关函数的Source (Py v2.7.3)非常有帮助。令人尴尬的是,我总是忘记Read the Source。我只是在搜索了类似的细节后才这么做的。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90+%到他们正在寻找的任何东西。沮丧的是,我想出了这个例子。我希望它能帮助其他人。(这对我很有帮助!;-)

代码语言:javascript
复制
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

在对这个问题的具体回答中:

代码语言:javascript
复制
sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno
票数 53
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/1278705

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