Question

Assume I have a program that uses argparse to process command line arguments/options. The following will print the ‘help’ message:

./myprogram -h

or:

./myprogram --help

But, if I run the script without any arguments whatsoever, it doesn’t do anything. What I want it to do is to display the usage message when it is called with no arguments. How is that done?

Answer 1

This answer comes from Steven Bethard on Google groups. I’m reposting it here to make it easier for people without a Google account to access.

You can override the default behavior of the error method:

import argparse
import sys

class MyParser(argparse.ArgumentParser):
    def error(self, message):
        sys.stderr.write('error: %s\n' % message)
        self.print_help()
        sys.exit(2)

parser = MyParser()
parser.add_argument('foo', nargs='+')
args = parser.parse_args()

---

Note that the above solution will print the help message whenever the error
method is triggered. For example, test.py --blah will print the help message
too if --blah isn’t a valid option.

If you want to print the help message only if no arguments are supplied on the
command line, then perhaps this is still the easiest way:

import argparse
import sys

parser=argparse.ArgumentParser()
parser.add_argument('foo', nargs='+')
if len(sys.argv)==1:
    parser.print_help(sys.stderr)
    sys.exit(1)
args=parser.parse_args()

---

Note that parser.print_help() prints to stdout by default. As init_js suggests, use parser.print_help(sys.stderr) to print to stderr.