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

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

import sys

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

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.

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.

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.

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

import sys

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