Set and require default Python script OptionParser

Question

The following "parser.add_option" statements work but if the script is run without an option/arg it will not complain. If an option/argument are not specified I would like it to display help (-h / --help) as default.

usage = "usage: %prog [options] arg"
parser = OptionParser(usage)
parser.add_option('-d', '--directory',
        action='store', dest='directory',
        default=None, help='specify directory')
parser.add_option('-f', '--file',
        action='store', dest='filename',
        default=None, help='specify file')
parser.add_option('-v', '--version',
                  action="store_true", dest="show_version",
                  default=False, help='displays the version number')
(options, args) = parser.parse_args()
#if len(args) < 1:
#    parser.error("incorrect number of arguments")

Secondly, if I enable the following snip than I get "error: incorrect number of arguments" even when specifying an option/arg.

if len(args) < 1:
parser.error("incorrect number of arguments")

Thanks.

Updated Code with Traceback error below

def main():
    usage = "usage: %prog [options] arg"
    parser = OptionParser(usage)
    parser.add_option('-d', '--directory',
            action='store', dest='directory',
            default=None, help='specify directory')
    parser.add_option('-f', '--file',
            action='store', dest='filename',
            default=None, help='specify file')
    parser.add_option('-v', '--version',
                      action="store_true", dest="show_version",
                      default=False, help='displays the version number')
    if len(sys.argv) == 1:
        parser.print_help()
        sys.exit()
    (options, args) = parser.parse_args()

#if options.show_version:
#    prog = os.path.basename(sys.argv[0])
#    version_str = "1.0"
#    print "version is: %s %s" % (prog, version_str)
#    sys.exit(0)

filenames_or_wildcards = []

# remove next line if you do not want allow to run the script without the -f -d
# option, but with arguments
filenames_or_wildcards = args # take all filenames passed in the command line

# if -f was specified add them (in current working directory)
if options.filename is not None:
    filenames_or_wildcards.append(options.filename)

Traceback

$ python boto-backup.py Traceback (most recent call last):   File "boto-backup.py", line 41, in <module>
    filenames_or_wildcards = args # take all filenames passed in the command line NameError: name 'args' is not defined

`len(args)` will always be at least one because `args[0]` is the name of the python script itself. Did you perhaps mean `if len(args)==1` — Matt Connolly, Mar 19 '11 at 01:56
@TheSoundOfMatt: you're thinking of `sys.argv`. In this case `len(args)` equals zero in the OP's scenario (of having no command-line input). — mechanical_meat, Mar 19 '11 at 02:09

mechanical_meat · Accepted Answer · 2011-03-19T02:37:49.287

I would do something like this:

from optparse import OptionParser
import sys

def main():
    usage = "usage: %prog [options] arg"
    parser = OptionParser(usage)
    parser.add_option('-d', '--directory',
            action='store', dest='directory',
            default=None, help='specify directory')
    parser.add_option('-f', '--file',
            action='store', dest='filename',
            default=None, help='specify file')
    parser.add_option('-v', '--version',
                      action="store_true", dest="show_version",
                      default=False, help='displays the version number')
    if len(sys.argv) == 1:
        parser.print_help()
        sys.exit()
    (options, args) = parser.parse_args()
    # rest of program...

if __name__ == '__main__':
    main()

So we set up the parser and options, and then check if there was any command-line input.
If none, we print the help message and exit. Otherwise we proceed with program execution.

When run with no command-line arguments the output is:

Usage: your_script_name_here.py [options] arg

Options:
  -h, --help            show this help message and exit
  -d DIRECTORY, --directory=DIRECTORY
                        specify directory
  -f FILENAME, --file=FILENAME
                        specify file
  -v, --version         displays the version number

Edit (in response to updated code):
Whitespace/indentation is important in Python.
Make sure that the rest of your code is indented such that it belongs to the main() function.
From filenames_or_wildcards = [] on, your code is outside of the scope of the main() function and thus doesn't have a variable named args.

@Astron: please edit your question to include the traceback. You are using `args` before it has been defined. — mechanical_meat, Mar 19 '11 at 02:20
is there an easy way to integrate the -d and -f? I have opened a new question: (http://stackoverflow.com/questions/5380279/combining-file-and-directory-selection-options) — Astron, Mar 21 '11 at 17:21

Set and require default Python script OptionParser

1 Answers1