passing arguments when starting shell script

Question

Hi I am trying to make a shell script to accept arguments at startup. The goal I want to achieve is to call a python script which by itself takes arguments. If the arguments are missing, default statement to make them predefined default values. This is the script I am trying to make work:

#!/bin/bash
#Script for calling th e python script sniff.py

while getopts d:D o
do case "$o" in
d)  usb_dev="@OPTARG";;
D)  console_outp="@OPTARG";;
*)  usb_dev=ttyUSB1;
    console_outp=INFO;      
esac
done

sniff=/home/user/Documents/Wireshark/Sensniff/sniff.py

sudo python $sniff -d "$usb_dev" -D "$console_outp"
sleep 5d

This script of course isn't working. Can you please help me to make this script working?

Answer 1

try

#!/bin/bash
#Script for calling th e python script sniff.py

usb_dev=ttyUSB1
console_outp=INFO

while getopts d:D: o
do case "$o" in
d)  usb_dev="$OPTARG";;
D)  console_outp="$OPTARG";;
esac
done

(same as your script)

• default value should be initialized before.
• use $OPTARG instead of @OPTARG
• option are defined as :D:d: