Question

我正在尝试创建一个通用错误处理程序，类似于Oracle中的“其他人”。我可以找到的示例都涉及捕获特定的预期错误。

Try:
    some_function()
Except: #I don't know what error I'm getting
    show_me_error(type_of_error_and_message)

Answer 1

try:
    1/0
except Exception as e:
    print '%s (%s)' % (e.message, type(e))

>>> 
integer division or modulo by zero (<type 'exceptions.ZeroDivisionError'>)

这非常well-documented。

但您可能只需要Sentry吗？

Answer 2

要使用Python 3打印异常，您将要使用type(e)。下面的示例：

try:
    1/0
except Exception as e:
    print(type(e))

>>> <class 'ZeroDivisionError'>

然后您可以使用以下命令捕获异常：

try:
    1/0
except ZeroDivisionError:
    print('Cannot divide by 0')
except Exception as e:
    print(type(e))

>>> Cannot divide by 0

Answer 3

import traceback

try:
    some_function()
except Exception as e:
    trace_back = traceback.format_exc()
    message = str(e)+ " " + str(trace_back)
    print message

Answer 4

要在Python 3中打印异常，

try:
    # your code
except Exception as e:
    print(e)

Django：捕获所有错误并显示错误和消息的类型

4 个答案: