Checking script's arguments in a function

Question

I am having difficulty trying to get an if statement working inside a function. I am using a very simple example like below to echo the $1 variable after script execution:

Logic inside script

STATUS() {

if [[ $1 == "-s" ]];then
   echo "--------------------------------------------------------"
   echo "Test success at: `date`:"
   echo "--------------------------------------------------------"
else
   echo "--------------------------------------------------------"
   echo "Test fail at: `date`:"
   echo "--------------------------------------------------------"
fi
}

STATUS

Example Execution:

ksh chris_test.sh -s

Expected Output:

--------------------------------------------------------
Test success at: Sun Jan  7 06:59:34 GMT 2024:
--------------------------------------------------------

What I am getting:

--------------------------------------------------------
Test fail at: Sun Jan  7 06:59:34 GMT 2024:
-------------------------------------------------------- 

I have attempted to escape the double quotes and the dollar sign, including the square brackets even trying to make square brackets single makes no difference. Still seeing fail. Where am I going wrong here?

Answer 1

Inside of STATUS(), $1 does not refer to the script's first argument but to the function's. When you call the function you'll need to forward the script's arguments so it can read them.

"$@" does that. It expands to "$1" "$2" "$3" ... for however many arguments there are, and with them properly quoted so spaces and other special characters are retained.

STATUS "$@"