How do I stop a program when an exception is raised in Python ?

Question

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

Answer 1

As far as I know, if an exception is not caught by your script, it will be interrupted.

Answer 2

import sys

try:
  print("stuff")
except:
  sys.exit(0)

Answer 3

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 4

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 anything to do to make your script exit when an exception happens.

Answer 5

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.