Passing arguments to Python script wrapped in shell script

Question

Consider the following Python script, script.py:

import argparse

parser = argparse.ArgumentParser()

parser.add_argument('-a', type=int)
parser.add_argument('-b', type=int)

args = parser.parse_args()

print('a + b = {}'.format(args.a + args.b))

And the following shell script, runner.sh:

python3 script.py

I know I can run script.py like this $ python3 script.py -a 4 -b 6 to get the following result: a + b = 10, but I want to run script.py from within the shell script, and be able to pass -a and -b, like this: ./runner.sh -a 4 -b 6. However, when I try this, those arguments are not passed:

$ ./runner.sh -a 10 -b 6
Traceback (most recent call last):
  File "script.py", line 10, in <module>
    print('a + b = {}'.format(args.a + args.b))
TypeError: unsupported operand type(s) for +: 'NoneType' and 'NoneType'

How can I pass arguments to a Python script that's wrapped inside a shell script? For what it's worth, I also tried sys.argv, to no avail.

runner.sh is literally python3 script.py in a .sh file, as shown above. — boot-scootin
– boot-scootin, Commented Aug 3, 2018 at 12:49

chepner · Accepted Answer · 2018-08-03 12:49:35Z

4

You need to tell the bash script to pass its arguments along to script.py:

#!/bin/bash

python script.py "$@"

Your script was ignoring its own arguments, and running script.py with no arguments.

answered Aug 3, 2018 at 12:49

chepner

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