when you want to pass multiple parameters through you shell script, you use $@
ex:

#! /bin/sh
echo $@

and then execute this file:

./script1 hello world

the script will prints “Hello world”

But be ware, $@ is not “$@” !!

The story is, I wrote some Java file that accepts 1 argument from the command line, such like this:

public class SayHello{
    public static void main(String[] args){
        if (args.length != 1){
            System.out.println("Usage: SayHello ");
            System.exit(-1);
        }
    // do other staff here
    }
}

I have wrapped this program in a shell file “sayHello.sh”:

#! /bin/sh
java SayHello $@

But the result was unexpected, as when I run with this command:

./sayHello.sh "Mohammed Hewedy"

I got

Usage: SayHello

I found the problem in the shell file, that I should use “$@” instead of $@
So, the script should be:

#! /bin/sh
java SayHello "$@"