我想达到这样的效果:
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')
但我得到的是:
Traceback..
IOError('Stuff')
关于如何实现这一点,有什么线索吗?如何在Python 2和3中都做到这一点?
我将提供一段我经常使用的代码片段,每当我想向异常添加额外信息时。我在Python 2.7和3.6中都可以工作。
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))
上面的代码产生如下输出:
---------------------------------------------------------------------------
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
我知道这与问题中提供的例子有一点偏差,但我希望有人觉得它有用。
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。
我使用的一个方便的方法是使用类属性作为详细信息的存储,因为类属性可以从类对象和类实例中访问:
class CustomError(Exception):
def __init__(self, details: Dict):
self.details = details
然后在代码中:
raise CustomError({'data': 5})
当捕获错误时:
except CustomError as e:
# Do whatever you want with the exception instance
print(e.details)