将信息添加到exception？
编辑：我正在运行Python 2.6
我想实现这样的事情：
def foo(): try: raise IOError('Stuff ') except: raise def bar(arg1): try: foo() except Exception as e: e.message = e.message + 'happens at %s' % arg1 raise bar('arg1') Traceback... IOError('Stuff Happens at arg1')
但是我得到的是：
Traceback.. IOError('Stuff')
任何线索，如何实现这一目标？
我会这样做，所以在foo()改变它的types也不需要在bar()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
这是一个稍微的修改，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.x，我的第一次更新中的代码在语法上是不正确的，加上在BaseException上拥有一个message属性的想法在BaseException被更改为PEP 352 （我的第一个更新发布在2012-03- 12）。 因此，目前在Python 3.5.2中，你需要沿着这些线做一些事情来保留回溯，而不是在函数bar()硬编码exception的types。 还要注意会有这样一行：
During handling of the above exception, another exception occurred:
在显示的回溯信息中。
# 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
一位评论者问，是否有一种方法可以在Python 2和Python 3中工作。虽然由于语法的不同，答案似乎是“否”，但是通过使用像reraise()这样的辅助函数可以解决这个问题。 six附加模块。 所以，如果你不想使用库，下面是一个简化的独立版本。
还要注意的是，由于在reraise reraise()函数内reraise()了exception，所以会出现在所有引发的回溯中，但最终的结果是你想要的。
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（），你可以试试这个：
try: raise IOError('stuff') except Exception as e: e.args = (e.args[0] + ' happens',) raise
如果你来这里寻找Python3 的解决scheme， 手册上说：
当引发一个新的exception（而不是使用裸
raise来重新引发当前正在处理的exception）时，隐式的exception上下文可以用一个明确的原因来补充：
raise new_exc from original_exc
例：
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
最后看起来像这样：
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.
把一个完全不重要的TypeError变成一个很好的消息，提示一个解决scheme，而不会搞乱原始exception。
你可以定义你自己的exceptioninheritance另一个exception，并创build它自己的构造函数来设置值。
例如：
class MyError(Exception): def __init__(self, value): self.value = value Exception.__init__(self) def __str__(self): return repr(self.value)
我使用的一个方便的方法是使用类属性作为存储的细节，因为类属性可以从类对象和类实例访问：
class CustomError(Exception): details = None
然后在你的代码中：
exc = CustomError('Some message') exc.details('Details -- add whatever you want') raise exc
当遇到一个错误：
except CustomError, e: # Do whatever you want with the exception instance print e print e.details
与以前的答案不同，这个工作面对的是非常糟糕的__str__exception。 但是，它会修改types，以便__str__无用的__str__实现。
我仍然希望find一个不会修改types的额外改进。
from contextlib import contextmanager @contextmanager def helpful_info(): try: yield except Exception as e: class CloneException(Exception): pass CloneException.__name__ = type(e).__name__ CloneException.__module___ = type(e).__module__ helpful_message = '%s\n\nhelpful info!' % e import sys raise CloneException, helpful_message, sys.exc_traceback class BadException(Exception): def __str__(self): return 'wat.' with helpful_info(): raise BadException('fooooo')
原始回溯和types（名称）被保留。
Traceback (most recent call last): File "re_raise.py", line 20, in <module> raise BadException('fooooo') File "/usr/lib64/python2.6/contextlib.py", line 34, in __exit__ self.gen.throw(type, value, traceback) File "re_raise.py", line 5, in helpful_info yield File "re_raise.py", line 20, in <module> raise BadException('fooooo') __main__.BadException: wat. helpful info!
我会提供一段代码，我经常使用，只要我想添加额外的信息到exception。 我在Python 2.7和3.6中工作。
import sys import traceback try: a = 1 b = 1j # The line below raises an exception because # we cannot compare int to complex. m = max(a, b) except Exception as ex: # I create my informational message for debugging: msg = "a=%r, b=%r" % (a, b) # Gather the information from the original exception: exc_type, exc_value, exc_traceback = sys.exc_info() # Format the original exception for a nice printout: traceback_string = ''.join(traceback.format_exception( exc_type, exc_value, exc_traceback)) # Re-raise a new exception of the same class as the original one, # using my custom message and the original traceback: raise type(ex)("%s\n\nORIGINAL TRACEBACK:\n\n%s\n" % (msg, traceback_string))
上面的代码产生以下输出：
--------------------------------------------------------------------------- TypeError Traceback (most recent call last) <ipython-input-6-09b74752c60d> in <module>() 14 raise type(ex)( 15 "%s\n\nORIGINAL TRACEBACK:\n\n%s\n" % ---> 16 (msg, traceback_string)) TypeError: a=1, b=1j ORIGINAL TRACEBACK: Traceback (most recent call last): File "<ipython-input-6-09b74752c60d>", line 7, in <module> m = max(a, b) # Cannot compare int to complex TypeError: no ordering relation is defined for complex numbers
我知道这个问题与这个问题提供的例子稍有不同，但我希望有人发现它有用。
也许
except Exception as e: raise IOError(e.message + 'happens at %s'%arg1)
