Linux

If you want true for empty strings, you want -z not -n

if [[ -z "VARIABLE1"  && -z "VARIABLE2"  ]]; then
    echo "OK"
fi

[ condition1 ] && [ condition2 ] && echo "Good" || echo "Bad"

To check for an empty string, use -z. -n checks to see if a string is not empty.

You're probably wanting [ -z "${VAR1}" -a -z "${VAR2}" ]. Note in bash that there are minor differences in how [ ] and [[ ]] tests are handled. You can pull up a handy cheat sheet of the operands on most distros by running man test, though you'll need to read through the CONDITIONAL EXPRESSIONS section of man bash if you want to see the minor differences of the single vs double square bracket commands (mostly whether locale applies to string order, as well as whether operands are evaluated in numeric comparisons).

You need to reference the value of the variable, ie:

if [[ -n "$VARIABLE1" && -n "$VARIABLE2" ]]; then
    echo "OK"
fi

Thank you all for yours input

What finally did work

if [[ -z VARIABLE1 && -z VARIABLE2 ]]; then
    echo "OK"
fi

If only Linux was using Python syntax that would be so much more intuitive...

Try this:

#!/usr/bin/env bash

a=""
if [[ -z "${a}" && -z "${b}" ]]; then
    echo "OK"
else
    echo "Not OK"
fi

a="OK"
if [[ -n "${a}" && -z "${b}" ]]; then
    echo "More ${a}"
else
    echo "More Unokay"
fi

Could try:

if [ condition1 ] && [ condition2 ]; then
  echo "OK"
fi