Test if multiple variables are set

Quoting error.

if [ -n "${!var}" ] ; then

For the future: Setting

set -x

before running the code would have shown you the problem. Instead of adding that to the code you can call your script with

bash -vx ./my/script.sh

Only thing you need are quotes in your test:

for var in one two three ; do
    if [ -n "${!var}" ] ; then
        echo "$var is set to ${!var}"
    else
        echo "$var is not set"
    fi
done

Works for me.

If you want the program stopped:

N= 
${one?var 1 is unset} 
${two:?var 2 is unset or null}
${three:+${N:?var 3 is set and not null}}

That'll do the trick. Each of the messages following the question mark is printed to stderr and the parent shell dies. Well, OK, so not each message - only one - just the first one that fails prints a message cause the shell dies. I like to use these tests like this:

( for v in "$one" "$two" "$three" ; do
    i=$((i+1)) ; : ${v:?var $i is unset or null...} 
done ) || _handle_it

I had a lot more to say about this here.