Re: How does one make argparse print usage when no options areprovided on the command line?
On Thu, 06 Dec 2012 00:06:29 -0500
Terry Reedy <email@example.com> wrote:
> On 12/5/2012 7:48 PM, rh wrote:
> > On Wed, 5 Dec 2012 18:42:37 +0100
> > Bruno Dupuis <firstname.lastname@example.org> wrote:
> >> On Wed, Dec 05, 2012 at 08:48:30AM -0800, rh wrote:
> >>> I have argparse working with one exception. I wanted the program
> >>> to print out usage when no command line options are given. But I
> >>> only came across other examples where people didn't use argparse
> >>> but instead printed out a separate usage statement. So they used
> >>> argparse for everything but the case where no command line args
> >>> are given.
> >> this is quite raw, but i'd add
> >> import sys
> >> if len(sys.argv) == 1:
> >> sys.argv.append('-h')
> > This works too. I guess I like the print_usage() method better.
> > Being new to python I have noticed that I had copied a bit of code
> > that did
> > if len(sys.argv[1:]) == 0:
> This needlessly creates and tosses a new object.
> > You did this:
> > if len(sys.argv) == 1:
> This does not.
> > The other reply did this:
> > if len(sys.argv) <= 1:
> This allows for the possibility that len(sys.argv) == 0. However,
> that can (according to the doc) only happen when starting the
> interpreter interactively without a script. Since that does not apply
> to code within a .py file, I prefer == 1.
> "argv is the script name (it is operating system dependent whether
> this is a full pathname or not). If the command was executed using
> the -c command line option to the interpreter, argv is set to the
> string '-c'. If no script name was passed to the Python interpreter,
> argv is the empty string."
Interesting, I thought maybe <= 1 was to handle some strange case
where sys.argv < 0. And I guess I stick with == 1. Always good to know
from the get go what's the best way to do the basics.
> Terry Jan Reedy
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