有没有办法将try / except块中的异常从一个传播到另一个除外?
我想捕获一个特定的错误,然后进行一般的错误处理.
“raise”允许异常“冒泡”到外部try / except,但不在try / except块内引发错误.
理想情况应该是这样的:
import logging
def getList():
try:
newList = ["just","some","place","holders"]
# Maybe from something like: newList = untrustedGetList()
# Faulty List now throws IndexError
someitem = newList[100]
return newList
except IndexError:
# For debugging purposes the content of newList should get logged.
logging.error("IndexError occured with newList containing: \n%s",str(newList))
except:
# General errors should be handled and include the IndexError as well!
logging.error("A general error occured,substituting newList with backup")
newList = ["We","can","work","with","this","backup"]
return newList
我遇到的问题是当IndexError被第一个捕获时除外,我的常规错误处理在第二个除了块之外没有应用.
我现在唯一的解决方法是在第一个块中包含一般错误处理代码.即使我将它包装在它自己的功能块中,它仍然看起来不那么优雅……
最佳答案
您有两种选择:
>不要使用专用的except块来捕获IndexError.您始终可以通过捕获BaseException并将异常分配给名称(此处为e)来手动测试常规块中的异常类型:
try:
# ...
except BaseException as e:
if isinstance(e,IndexError):
logging.error("IndexError occured with newList containing: \n%s",str(newList))
logging.error("A general error occured,substituting newList with backup")
newList = ["We","backup"]
return newList
>使用嵌套的try..except语句并重新引发:
try:
try:
# ...
except IndexError:
logging.error("IndexError occured with newList containing: \n%s",str(newList))
raise
except:
logging.error("A general error occured,"backup"]
return newList
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