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 set "required=1" but I got an error:

invalid keyword arguments: required

I want to make my script require --file option to be input by users. I know that the action keyword gives you error when you don't supply value to --file whose action="store_true".

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') 

Answer 2

Since if not x doesn't work for some(negative,zero) parameters,

and to prevent lots of if tests, i preferr something like this:

required="host username password".split()  parser = OptionParser() parser.add_option("-H", '--host', dest='host') parser.add_option("-U", '--user', dest='username') parser.add_option("-P", '--pass', dest='password') parser.add_option("-s", '--ssl',  dest='ssl',help="optional usage of ssl")  (options, args) = parser.parse_args()  for r in required:     if options.__dict__[r] is None:         parser.error("parameter %s required"%r)