Some_function()在执行时引发异常,因此程序跳转到异常:

try:
    some_function()
except:
    print("exception happened!")

如何查看导致异常发生的原因?


当前回答

希望这对你有帮助

import sys
varExcepHandling, varExcepHandlingZer = 2, 0
try:
  print(varExcepHandling/varExcepHandlingZer)
except Exception as ex: 
  print(sys.exc_info())

'sys.exc_info()'将返回一个元组,如果你只想要异常类名,请使用'sys.exc_info()[0]'

注意:-如果你想看到所有的异常类,只需写入dir(__builtin__)

其他回答

你通常不应该用try来捕获所有可能的异常:…除非这太宽泛了。只要抓住那些无论出于什么原因都会发生的事情。如果您真的必须这样做,例如,如果您想在调试时了解有关某个问题的更多信息,那么您应该这样做

try:
    ...
except Exception as ex:
    print ex # do whatever you want for debugging.
    raise    # re-raise exception.

大多数答案指向except(…)as(…):语法(正确),但同时没有人想谈论房间里的大象,而大象是sys.exc_info()函数。 来自sys模块的文档(重点是我的):

This function returns a tuple of three values that give information about the exception that is currently being handled. (…) If no exception is being handled anywhere on the stack, a tuple containing three None values is returned. Otherwise, the values returned are (type, value, traceback). Their meaning is: type gets the type of the exception being handled (a subclass of BaseException); value gets the exception instance (an instance of the exception type); traceback gets a traceback object (see the Reference Manual) which encapsulates the call stack at the point where the exception originally occurred.

我认为sys.exc_info()可以被视为对原始问题“我如何知道发生了什么类型的异常?”的最直接答案。

希望这对你有帮助

import sys
varExcepHandling, varExcepHandlingZer = 2, 0
try:
  print(varExcepHandling/varExcepHandlingZer)
except Exception as ex: 
  print(sys.exc_info())

'sys.exc_info()'将返回一个元组,如果你只想要异常类名,请使用'sys.exc_info()[0]'

注意:-如果你想看到所有的异常类,只需写入dir(__builtin__)

使用类型class和as语句

try:#code
except Exception as e:
     m=type(e)
     #m is the class of the exception
     strm=str(m)
     #strm is the string of m

你可以像Lauritz推荐的那样开始:

except Exception as ex:

然后像这样打印ex:

try:
    #your try code here
except Exception as ex:
    print ex