[SOLVED] Can I only call a function from the terminal rather the whole bash script?

Issue

I have a bash script that looks as such:

#!/bin/bash

function one {
    echo "I am function one!!"
}

function two {
    echo "I am function two!!"
}

one
two

If I simply do bash test.sh both functions are being executed.

What I'd like to do is to call the script from the terminal while also specifying one of the two functions, and executing only it. Maybe something like: bash test.sh$one() and it should only print out

I am function one!!

Is this possible and if so, how will I go about achieving it?

Thanks!

=========================

EDIT: As per @Waqas suggestion I ended up implementing the below which did the trick for me:

function main {
    if [ -z "$1" ]
    then
        some commands
    # else run the given function only
    else
        $1
    fi
}
main "$@"

Thanks!!!

Solution

There are many ways to write the code in order to fulfill your requirement. The way I will write the code for this, is the following:

#!/bin/bash
    
function main {
    # If the argument is empty then run both functions else only run provided function as argument $1.
    [ -z "$1" ] && { one; two; } || $1     
}
        
function one {    
    echo "I am function one!!"    
}
    
function two {    
    echo "I am function two!!"    
}
    
main "$@"

If you only execute the script without passing argument then both functions will run and with passing argument only single function will work.

Example1 (Both functions will run): bash script_name

Example2 (Only function one will run): bash script_name one

Example3 (Only function two will run): bash script_name two

Answered By - Waqas Naim Khan