Question

I want to exit a Python script if my try succeeds. My code just doesn’t exit and continues to execute the rest of the script.

If I run python myscript.py --help, I need the script to exit. If I run python myscript.py I need do execute the except part.

try:
    if sys.argv[1] == '--help' or sys.argv[1] == '-help' or \
            sys.argv[1] == '-h' or sys.argv[1] == '--h':
        print "NAME"
        sys.exit("Usage : python go.py")
except:
    #REST OF THE CODE.

Answer 1

Why would you use a try / except block? Like bereal says, you are catching all exceptions and hiding problems. try /except blocks are for exceptional circumstances. Not alternate code paths.

def main():
    # main code block

if sys.argv[1] == '--help' or sys.argv[1] == '-help' or \
    sys.argv[1] == '-h' or sys.argv[1] == '--h':
    print "Help text"
else:
    main()

Also, as a general rule when using try /except. Just catch the exception that you care about. It’s impossible to debug something that cannot even tell you what is wrong.