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