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
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
.