str_replace a NULL value - php

Question

I'm trying to use str_replace to replace '' values to 'phone', I've used the code below but it isn't replacing anything:
<?php
$start_up = str_replace('','phone','start_up');
$sales = str_replace('','phone','');
$resourcing = str_replace('','phone','resourcing');
$management = str_replace('','phone','');
$array = "($start_up OR $sales OR $resourcing OR $management)";
echo $array;
?>
I want this to come up with:
(start_up OR phone OR resourcing OR phone)
But instead it is doing this:
(start_up OR OR resourcing OR )
I know I can use a !isset($var) type query but this seems clunky and long winded. Is there a way to include blank strings in a str_replace query?

Answer 1

You can use preg_replace instead of str_replace:
$start_up = preg_replace('/^$/','phone','start_up');
$sales = preg_replace('/^$/', 'phone', '');
$resourcing = preg_replace('/^$/','phone','resourcing');
$management = preg_replace('/^$/','phone','');
$array = "($start_up OR $sales OR $resourcing OR $management)";
echo $array;
Output:
(start_up OR phone OR resourcing OR phone)
and it will replace, an empty string with phone.
Edit: OR you can use empty() and ternary operator:
$start_up = empty('start_up') ? 'phone' : 'start_up';
$sales = empty('') ? 'phone' : '';
$resourcing = empty('resourcing') ?'phone' : 'resourcing';
$management = empty('') ? 'phone' : '';
$array = "($start_up OR $sales OR $resourcing OR $management)";
echo $array;
Output same as above.