Understanding boolean operators in bash script

The variable $phone_missing is a string that happens to contain false. And a non-empty string evaluates to true. See also http://www.linuxintro.org/wiki/Babe#empty_strings

I often use "true" and "false" since they are also commands that merely return success and failure respectively. Then you can do

if "$phone_missing"; then ...

Here's one way to do this, while retaining the true/false values.

phone_missing=false
if [ "$phone_missing" != false ]; then
    echo "phone_missing is not 'false' (but may be non-true, too)"
fi
if [ "$phone_missing" == true ]; then
    echo "phone_missing is true."
fi

The double quotes around $phone_missing are to protect against the case where variable phone_missing is not defined at all. Another common idiom to ward against this is [ x$phone_missing != xfalse ], but the quotes seem more natural to me.

The hint is in the bash help page for test:

  STRING      True if string is not empty.
  ...
  ! EXPR      True if expr is false.

So, basically [ $foo ] will be true if $foo is non-empty. Not true or false, just non-empty. [ ! $foo ] is true if $foo is empty or undefined.

You could always change your code to just set phone_missing to a non-empty value, which will denote true. If phone_missing is unset (or empty — phone_missing=""), it will be false. Otherwise, you should be using the string testing operators (= and !=).

The other slight issue is the assignment. You have it as $phone_missing=true, whereas it should be phone_missing=true (no dollar sign).

Sorry if this is a bit dense, it's because I am. It's been a long day. :)