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。
到目前为止,我对所有给出的答案都不满意。他们还是太啰嗦了。在代码和消息输出中。
所有我想要的是stacktrace指向源异常,没有异常之间的东西,所以没有创建新的异常,只是重新提升原始与所有相关的堆栈帧状态在它,导致那里。
史蒂夫·霍华德给出了一个很好的答案,我想扩展一下,不,是简化一下……仅适用于python 3。
except Exception as e:
e.args = ("Some failure state", *e.args)
raise
唯一的新东西是参数展开/解包,这使得它对我来说足够小和容易使用。
试一试:
foo = None
try:
try:
state = "bar"
foo.append(state)
except Exception as e:
e.args = ("Appending '"+state+"' failed", *e.args)
raise
print(foo[0]) # would raise too
except Exception as e:
e.args = ("print(foo) failed: " + str(foo), *e.args)
raise
这将给你:
Traceback (most recent call last):
File "test.py", line 6, in <module>
foo.append(state)
AttributeError: ('print(foo) failed: None', "Appending 'bar' failed", "'NoneType' object has no attribute 'append'")
简单的漂亮图案可以是这样的
print("\n".join( "-"*i+" "+j for i,j in enumerate(e.args)))