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?
当前回答
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
其他回答
与之前的答案不同，这适用于非常糟糕的__str__异常。 但是，它确实修改了类型，以便剔除无用的__str__实现。
我仍然希望找到一个不修改类型的额外改进。
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')
原始的回溯和类型(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!
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)
这是我的实现，使用它作为上下文管理器，并可选地添加额外的消息异常:
from typing import Optional, Type
from types import TracebackType
class _addInfoOnException():
def __init__(self, info: str = ""):
self.info = info
def __enter__(self):
return
def __exit__(self,
exc_type: Optional[Type[BaseException]],
exc_val: BaseException, # Optional, but not None if exc_type is not None
exc_tb: TracebackType): # Optional, but not None if exc_type is not None
if exc_type:
if self.info:
newMsg = f"{self.info}\n\tLow level error: "
if len(exc_val.args) == 0:
exc_val.args = (self.info, )
elif len(exc_val.args) == 1:
exc_val.args = (f"{newMsg}{exc_val.args[0]}", )
elif len(exc_val.args) > 0:
exc_val.args = (f"{newMsg}{exc_val.args[0]}", exc_val.args[1:])
raise
用法:
def main():
try:
raise Exception("Example exception msg")
except Exception:
traceback.print_exc()
print("\n\n")
try:
with _addInfoOnException():
raise Exception("Example exception msg, no extra info")
except Exception:
traceback.print_exc()
print("\n\n")
try:
with _addInfoOnException("Some extra info!"):
raise Exception("Example exception msg")
except Exception:
traceback.print_exc()
print("\n\n")
if __name__ == "__main__":
main()
这将在这样的跟踪中解决:
Traceback (most recent call last):
File "<...>\VSCodeDevWorkspace\testis.py", line 40, in main
raise Exception("Example exception msg")
Exception: Example exception msg
Traceback (most recent call last):
File "<...>\VSCodeDevWorkspace\testis.py", line 47, in main
raise Exception("Example exception msg, no extra info")
File "<...>\VSCodeDevWorkspace\testis.py", line 47, in main
raise Exception("Example exception msg, no extra info")
Exception: Example exception msg, no extra info
Traceback (most recent call last):
File "<...>\VSCodeDevWorkspace\testis.py", line 54, in main
raise Exception("Example exception msg")
File "<...>\VSCodeDevWorkspace\testis.py", line 54, in main
raise Exception("Example exception msg")
Exception: Some extra info!
Low level error: Example exception msg
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')
