substituting a string in place of variable in shell

Question

I pass a string as an argument to a shell script. and the shell script should tell me if the passed argument is a variable

something like this

if [ ! -z          


        

Answer 1

The syntax for this is:

${!VAR}

Example:

$ function hello() { echo ${!1}; }
$ hello HOME
/home/me

Answer 2

Found it here: http://www.linuxquestions.org/questions/programming-9/bash-how-to-get-variable-name-from-variable-274718/

All you should do:

if [ ! -z ${!1} ]; then
    echo yes $1 is a variable and its value is ${!1}
fi