在 Python 中引发异常时如何停止程序？

Question

I need to stop my program when an exception is raised in Python. How do I implement this?

Answer 1

import sys

try:
  print("stuff")
except:
  sys.exit(1) # exiing with a non zero value is better for returning from an error

Answer 2

You can stop catching the exception, or - if you need to catch it (to do some custom handling), you can re-raise:

try:
  doSomeEvilThing()
except Exception, e:
  handleException(e)
  raise

Note that typing raise without passing an exception object causes the original traceback to be preserved. Typically it is much better than raise e .

Of course - you can also explicitly call

import sys 
sys.exit(exitCodeYouFindAppropriate)

This causes SystemExit exception to be raised, and (unless you catch it somewhere) terminates your application with specified exit code.

Answer 3

If you don't handle an exception, it will propagate up the call stack up to the interpreter, which will then display a traceback and exit. IOW : you don't have to do anything to make your script exit when an exception happens.

Answer 4

import sys

try:
    import feedparser
except:
    print "Error: Cannot import feedparser.\n" 
    sys.exit(1)

Here we're exiting with a status code of 1. It is usually also helpful to output an error message, write to a log, and clean up.

Answer 5

据我所知，如果您的脚本没有捕获异常，它将被中断。

Answer 6

import sys

try: 
    # your code here
except Exception as err:
    print("Error: " + str(err))
sys.exit(50) # whatever non zero exit code