How to exit from Python without traceback?


Question

I would like to know how to I exit from Python without having an traceback dump on the output.

I still want want to be able to return an error code but I do not want to display the traceback log.

I want to be able to exit using exit(number) without trace but in case of an Exception (not an exit) I want the trace.

1
264
1/26/2015 8:25:48 PM

Accepted Answer

You are presumably encountering an exception and the program is exiting because of this (with a traceback). The first thing to do therefore is to catch that exception, before exiting cleanly (maybe with a message, example given).

Try something like this in your main routine:

import sys, traceback

def main():
    try:
        do main program stuff here
        ....
    except KeyboardInterrupt:
        print "Shutdown requested...exiting"
    except Exception:
        traceback.print_exc(file=sys.stdout)
    sys.exit(0)

if __name__ == "__main__":
    main()
283
6/30/2012 7:48:53 PM

Perhaps you're trying to catch all exceptions and this is catching the SystemExit exception raised by sys.exit()?

import sys

try:
    sys.exit(1) # Or something that calls sys.exit()
except SystemExit as e:
    sys.exit(e)
except:
    # Cleanup and reraise. This will print a backtrace.
    # (Insert your cleanup code here.)
    raise

In general, using except: without naming an exception is a bad idea. You'll catch all kinds of stuff you don't want to catch -- like SystemExit -- and it can also mask your own programming errors. My example above is silly, unless you're doing something in terms of cleanup. You could replace it with:

import sys
sys.exit(1) # Or something that calls sys.exit().

If you need to exit without raising SystemExit:

import os
os._exit(1)

I do this, in code that runs under unittest and calls fork(). Unittest gets when the forked process raises SystemExit. This is definitely a corner case!


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