Print an error message without printing a traceback and close the program when a condition is not met

You can turn off the traceback by limiting its depth.

Python 2.x

import syssys.tracebacklimit = 0

Python 3.x

In Python 3.5.2 and 3.6.1, setting tracebacklimit to 0 does not seem to have the intended effect. This is a known bug. Note that -1 doesn't work either. Setting it to None does however seem to work, at least for now.

>>> import sys>>> sys.tracebacklimit = 0>>> raise ExceptionTraceback (most recent call last):  File "<stdin>", line 1, in <module>Exception>>> sys.tracebacklimit = -1>>> raise ExceptionTraceback (most recent call last):  File "<stdin>", line 1, in <module>Exception>>> sys.tracebacklimit = None>>> raise ExceptionException

Nevertheless, for better or worse, if multiple exceptions are raised, they can all still be printed. For example:

socket.gaierror: [Errno -2] Name or service not knownDuring handling of the above exception, another exception occurred:urllib.error.URLError: <urlopen error [Errno -2] Name or service not known>

You can use SystemExit exception:

except Exception as err:    raise SystemExit(err)

https://docs.python.org/3/library/exceptions.html

You can use a try: and then except Exception as inst:What that will do is give you your error message in a variable named inst and you can print out the arguments on the error with inst.args. Try printing it out and seeing what happens, and is any item in inst.args is the one you are looking for.

EDIT Here is an example I tried with pythons IDLE:

>>> try:    open("epik.sjj")except Exception as inst:    d = inst>>> dFileNotFoundError(2, 'No such file or directory')>>> d.args(2, 'No such file or directory')>>> d.args[1]'No such file or directory'>>> 

EDIT 2: as for closing the program you can always raise and error or you can use sys.exit()