Bash scripting - adding $ to make a variable

Question

Hi,
I'm trying to code the following, which should look like:
test(){
   if [ ! -z "$PATH" ]; then
      echo PATH=$PATH
   fi
test... but haven't been successful so far in in trying to pass "PATH" as a variable to the function.
For instance, just do illustrate what I mean:
test() {
   if [ ! -z "$[$1]" ]; then
      echo $1=$[$1]
   fi
test PATHMaybe this cannot be done?
I was also trying the following, but how to print the word PATH or $PATH
in the function, instead of printing the content of $1 ?
test(){
   if [ ! -z "$1" ]; then
      echo PATH=$1
   fi
test $PATHI guess I could do the following, but perhaps there is another way?
test(){
   if [ ! -z "$1" ]; then
      echo $2=$1
   fi
test $PATH PATHI was also wondering about the "-z" if $1 resolves in a string that is too long. I just want to check if variable $PATH exists.
Any ideas?
Thanks!
Edited by: Dude on Feb 14, 2011 7:36 PM

Answer 1

Thanks very much!
eval is the key. But in order to get what I wanted, I had to change it somewhat. Below functions finally produce the same output. (Thanks to you, I used eval as a search term in Google and found a similar example).
howto()     {
   if eval [ ! -z "\$$1" ]; then
     eval echo $1=\$$1
   fi
howto PATH
howto2() {
   if [ ! -z "$PATH" ]; then
      echo PATH=$PATH
   fi
howto2Edited by: Dude on Feb 15, 2011 6:10 AM