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Re: How does one make argparse print usage when no options are provided on the command line?

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  Re: How does one make argparse print usage when no options are provided on the command line? Terry Reedy <tjreedy@udel.edu> - 2012-12-06 00:06 -0500

#34361 — Re: How does one make argparse print usage when no options are provided on the command line?

On 12/5/2012 7:48 PM, rh wrote:
> On Wed, 5 Dec 2012 18:42:37 +0100
> Bruno Dupuis <python.ml.bruno.dupuis@lisael.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[0] 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[0] is set to the string 
'-c'. If no script name was passed to the Python interpreter, argv[0] is 
the empty string."

-- 
Terry Jan Reedy

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