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?
Advertisement
Answer
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: %sn' % 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.