I use the following simple code to parse some arguments; note that one of them is required. Unfortunately, when the user runs the script without providing the argument, the displayed usage/help text does not indicate that there is a non-optional argument, which I find very confusing. How can I get python to indicate that an argument is not optional?
Here is the code:
import argparse
if __name__ == '__main__':
parser = argparse.ArgumentParser(
description='Foo')
parser.add_argument('-i','--input', help='Input file name', required=True)
parser.add_argument('-o','--output', help='Output file name', default="stdout")
args = parser.parse_args()
print ("Input file: %s" % args.input )
print ("Output file: %s" % args.output )
When running above code without providing the required argument, I get the following output:
usage: foo.py [-h] -i INPUT [-o OUTPUT]
Foo
optional arguments:
-h, --help show this help message and exit
-i INPUT, --input INPUT
Input file name
-o OUTPUT, --output OUTPUT
Output file name
-i INPUT
part is not surrounded by square brackets, which subtlety indicates that is indeed, required. Also, you can manually explain that through thehelp
paramoptional arguments
for the required arguments is still misleading.