我在代码中使用:

try:
    a=1
    b=0
    c=a/b

except:
    raise Exception(f"can't divide {a} with {b}")

输出:

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_11708/1469673756.py in <module>
      3     b=0
----> 4     c=a/b
      5 

ZeroDivisionError: division by zero

During handling of the above exception, another exception occurred:

Exception                                 Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_11708/1469673756.py in <module>
      5 
      6 except Exception:
----> 7     raise Exception(f"can't divide {a} with {b}")

Exception: can't divide 1 with 0

我在代码中使用:

try:
    a=1
    b=0
    c=a/b

except:
    raise Exception(f"can't divide {a} with {b}")

输出:

---------------------------------------------------------------------------
ZeroDivisionError                         Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_11708/1469673756.py in <module>
      3     b=0
----> 4     c=a/b
      5 

ZeroDivisionError: division by zero

During handling of the above exception, another exception occurred:

Exception                                 Traceback (most recent call last)
~\AppData\Local\Temp/ipykernel_11708/1469673756.py in <module>
      5 
      6 except Exception:
----> 7     raise Exception(f"can't divide {a} with {b}")

Exception: can't divide 1 with 0

如果你来这里寻找Python 3的解决方案，手册上说:

当引发一个新的异常时(而不是使用一个简单的raise来重新引发当前正在处理的异常)，隐式异常上下文可以通过使用from和raise来补充显式原因:

raise new_exc from original_exc

例子:

try:
    return [permission() for permission in self.permission_classes]
except TypeError as e:
    raise TypeError("Make sure your view's 'permission_classes' are iterable. "
                    "If you use '()' to generate a set with a single element "
                    "make sure that there is a comma behind the one (element,).") from e

最后是这样的:

2017-09-06 16:50:14,797 [ERROR] django.request: Internal Server Error: /v1/sendEmail/
Traceback (most recent call last):
File "venv/lib/python3.4/site-packages/rest_framework/views.py", line 275, in get_permissions
    return [permission() for permission in self.permission_classes]
TypeError: 'type' object is not iterable 

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
    # Traceback removed...
TypeError: Make sure your view's Permission_classes are iterable. If 
     you use parens () to generate a set with a single element make 
     sure that there is a (comma,) behind the one element.

将一个完全无描述的TypeError转换为一个带有解决方案提示的漂亮消息，而不会搞乱原始异常。

我会这样做，这样在foo()中改变它的类型就不需要在bar()中也改变它。

def foo():
    try:
        raise IOError('Stuff')
    except:
        raise

def bar(arg1):
    try:
        foo()
    except Exception as e:
        raise type(e)(e.message + ' happens at %s' % arg1)

bar('arg1')

Traceback (most recent call last):
  File "test.py", line 13, in <module>
    bar('arg1')
  File "test.py", line 11, in bar
    raise type(e)(e.message + ' happens at %s' % arg1)
IOError: Stuff happens at arg1

更新1

这里有一个轻微的修改，保留了原始的回溯:

...
def bar(arg1):
    try:
        foo()
    except Exception as e:
        import sys
        raise type(e), type(e)(e.message +
                               ' happens at %s' % arg1), sys.exc_info()[2]

bar('arg1')

Traceback (most recent call last):
  File "test.py", line 16, in <module>
    bar('arg1')
  File "test.py", line 11, in bar
    foo()
  File "test.py", line 5, in foo
    raise IOError('Stuff')
IOError: Stuff happens at arg1

更新2

对于Python 3。在2012-05-16对PEP 352的修改中(我的第一次更新发布于2012-03-12)，我的第一次更新中的代码在语法上是错误的，加上在BaseException上有消息属性的想法被撤销了。因此，目前在Python 3.5.2中，您需要按照这些行做一些事情来保留回溯，而不是在函数栏()中硬编码异常的类型。还要注意，会有这样一行:

During handling of the above exception, another exception occurred:

在显示的回溯消息中。

# for Python 3.x
...
def bar(arg1):
    try:
        foo()
    except Exception as e:
        import sys
        raise type(e)(str(e) +
                      ' happens at %s' % arg1).with_traceback(sys.exc_info()[2])

bar('arg1')

更新3

一位评论者询问是否有一种方法可以同时在Python 2和3中工作。尽管由于语法差异，答案似乎是“否”，但有一种解决方法，即使用six外接组件模块中的rerraise()等帮助函数。因此，如果您出于某种原因不想使用这个库，下面是一个简化的独立版本。

还要注意，由于异常是在rerraise()函数中重新引发的，因此无论引发什么跟踪，都会出现异常，但最终结果是您想要的结果。

import sys

if sys.version_info.major < 3:  # Python 2?
    # Using exec avoids a SyntaxError in Python 3.
    exec("""def reraise(exc_type, exc_value, exc_traceback=None):
                raise exc_type, exc_value, exc_traceback""")
else:
    def reraise(exc_type, exc_value, exc_traceback=None):
        if exc_value is None:
            exc_value = exc_type()
        if exc_value.__traceback__ is not exc_traceback:
            raise exc_value.with_traceback(exc_traceback)
        raise exc_value

def foo():
    try:
        raise IOError('Stuff')
    except:
        raise

def bar(arg1):
    try:
       foo()
    except Exception as e:
        reraise(type(e), type(e)(str(e) +
                                 ' happens at %s' % arg1), sys.exc_info()[2])

bar('arg1')

假设你不想或不能修改foo()，你可以这样做:

try:
    raise IOError('stuff')
except Exception as e:
    if len(e.args) >= 1:
        e.args = (e.args[0] + ' happens',) + e.args[1:]
    raise

这确实是在Python 3中解决问题的唯一解决方案，而不会出现丑陋而令人困惑的“在处理上述异常期间，发生了另一个异常”消息。

如果要将重新提升的行添加到堆栈跟踪中，则应该写入raise e而不是raise。

以下是我在个人项目中使用的方法(我相信有足够的理由不在产品代码中这样做):

try:
    #something hazardous
except Exception as e:
    insightful_message = "shouldn't have done that"
    amended_args = tuple([f'{e.args[0]}\n{insightful_message}', *e.args[1:]])
    e.args = amended_args
    raise

代码(1)拦截错误;(2)创建错误的.args属性的副本，这是一个元组，假定在索引0处包含一个错误消息，使用列表理解实现;(3)在错误消息后追加换行符和自定义消息;(4)使用。args附加到副本的任何附加项 拆包;(5)将副本转换为元组;最后(6)用修改后的副本替换。args。

这些操作大多是为了避免.args元组的不可变性。