[SOLVED] Passing variable number of arguments to a shell script

Issue

I am building a script (linux bash) where I need to pass a certain number of arguments, it can be 7 or 8 arguments.

If I pass 7 arguments I will do it like this:

my_script.sh

!/bin/bash

arg1=$1
arg2=$2
arg3=$3
arg4=$4
arg5=$5
arg6=$6
arg7=$7
#Do other stuff after this

and I run it like this: ./my_script.sh 1 2 3 4 5 6 7

I want to be able to add an optional 8th parameter into this script. So the idea is to run the script with this 8th parametes sometimes and the rest of the time only with 7 parameters. How can I do this?

Solution

You could do something like this:

if [[ $8 != '' ]]
then
  # do something
fi

Also, if you do

arg8=$8

and if there is no eighth arg, arg8 will just remain empty.

Answered By - develc

Answer Checked By - Willingham (WPSolving Volunteer)