当我捕获异常时，如何获取类型、文件和行号？

问题描述：

捕获异常并打印如下内容：

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：

ZeroDivisonError, tmp.py, 1

解决方案 1：

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)

解决方案 2：

对我来说最简单的形式。

import traceback

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

输出

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

解决方案 3：

traceback.format_exception()和调用/相关函数的源代码(Py v2.7.3)非常有用。令人尴尬的是，我总是忘记阅读源代码。我徒劳地搜索了类似的细节后才这样做。一个简单的问题，“如何为异常重新创建与 Python 相同的输出，并包含所有相同的细节？”这会让任何人获得他们所寻找的 90% 以上的信息。沮丧之下，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
''' # 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

对这个问题的具体回答是：

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

解决方案 4：

下面是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')

解决方案 5：

无需导入回溯即可实现此目的：

try:
    func1()
except Exception as ex:
    trace = []
    tb = ex.__traceback__
    while tb is not None:
        trace.append({
            "filename": tb.tb_frame.f_code.co_filename,
            "name": tb.tb_frame.f_code.co_name,
            "lineno": tb.tb_lineno
        })
        tb = tb.tb_next
    print(str({
        'type': type(ex).__name__,
        'message': str(ex),
        'trace': trace
    }))

输出：

{

  'type': 'ZeroDivisionError',
  'message': 'division by zero',
  'trace': [
    {
      'filename': '/var/playground/main.py',
      'name': '<module>',
      'lineno': 16
    },
    {
      'filename': '/var/playground/main.py',
      'name': 'func1',
      'lineno': 11
    },
    {
      'filename': '/var/playground/main.py',
      'name': 'func2',
      'lineno': 7
    },
    {
      'filename': '/var/playground/my.py',
      'name': 'test',
      'lineno': 2
    }
  ]
}

解决方案 6：

无需任何导入，但也不会递归到导入的模块：

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 2

重申一下，这不适用于imports 或模块，因此如果这样做import X; try: X.example();，文件名和行号将指向包含的行，而不是在 中X.example()出错的行。如果有人知道如何从最后一个堆栈跟踪行轻松获取文件名和行号（我期望类似的东西，但没有那么幸运），请改进这个答案。 X.example()`e[-1].filename`

解决方案 7：

try:
    4/0
except Exception as exc:
    print("error: ",exc)
    print("error file info: ",exc.__traceback__.tb_frame)
    print("error line#: ",exc.__traceback__.tb_lineno)

解决方案 8：

这就是我用来获取文件名的方法。

__file__.__str__

总而言之，我创建了一个页面来显示错误。出现异常时返回此页面。

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

发生异常时的情况如下：