Python: How to make an option to be required in optparse?

Question

I\'ve read this http://docs.python.org/release/2.6.2/library/optparse.html

But I\'m not so clear how to make an option to be required in optparse?

I\'ve tried to

Answer 1

You can implement a required option easily.

parser = OptionParser(usage='usage: %prog [options] arguments')
parser.add_option('-f', '--file', 
                        dest='filename',
                        help='foo help')
(options, args) = parser.parse_args()
if not options.filename:   # if filename is not given
    parser.error('Filename not given')