python argparse: How can I display help automatically on error?

Question

Currently when I enter invalid options or omit positional arguments, argparse kicks me back to the prompt and displays the usage for my app. This is ok, but I would rather automatically display the full help listing (that explains the options, etc) than require the user to type

./myscript.py -h

Thanks!

Jamie

Asked By: jpswain.w

||

Source

Answer 1

This thread over at Google groups has the following code snippet which seems to do the trick (modified slightly).

class DefaultHelpParser(argparse.ArgumentParser):
    def error(self, message):
        sys.stderr.write('error: %sn' % message)
        self.print_help()
        sys.exit(2)

Answered By: user201788

Answer 2

To print help you might want to use: print_help function on ArgumentParser instance

parser = argparse.ArgumentParser()
(...)
parser.print_help()

To print help message on error you need to create own subclass of ArgumentParser instance, that overrides error() method. For example like that:

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

When this parser encounters unparseable argument line it will print help.

Answered By: jb.

Answer 3

Suppress printing of usage with usage=argparse.SUPPRESS. Then catch the SystemExit exception that ArgumentParser raises on error, print the help, and exit by raising the exception again.

parser = argparse.ArgumentParser(usage=argparse.SUPPRESS)
parser.add_argument(...)
try:
    args = parser.parse_args()
except SystemExit:
    parser.print_help()
    raise

Answered By: Roger Dahl

Answer 4

You, also, can print help without using a class or exception:

def _error(parser):
    def wrapper(interceptor):
        parser.print_help()

        sys.exit(-1)

    return wrapper

def _args_get(args=sys.argv[1:]):
    parser = argparser.ArgumentParser()

    parser.error = _error(parser)

    parser.add_argument(...)
    ...

. Just wrap ArgumentParser.error function in your and intercept message argument. I answered, there, earlier:

https://stackoverflow.com/a/60714163/10152015

Answered By: hesed

Answer 5

I just fixed this same problem myself using the following syntax:

parser = ArgumentParser()
... add arguments ...
parser.usage = parser.format_help()
args = parser.parse_args()

Answered By: infinitesteps

python argparse: How can I display help automatically on error?

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