preg_replace replace only once even if the match is found - php

Question

my HTML form code replaces some words with <-#word#-> using the code
$string = preg_replace("/($p)/i", '<-#$1#->', $string);
the problem is that if the form has some errors, upon resubmitting the form the word becomes <-#<-#<-#word#->#->#-> every time someone resubmits the form. Is it possible to replace but if it is already replaced then do not.
This is what I tried using NOT operator but it is not working
$string = preg_replace("/^(<-#)($p)^(#->)/i", '<-#$1#->', $string);

Answer 1

You could use a negative lookarounds to assert what is directly on the left an on the right is not <-# and
(?<!<-#)(word)(?!#->)
Regex demo | Php demo
Your code could look like:
$string = preg_replace("/(?<!<-#)($p)(?!#->)/i", '<-#$1#->', $string);

Answer 2

Another method might be to check with preg_match_all() to ensure if your matches are returning:
$string = '<-#<-#<-#Any alphanumeric input that user may wish#->#->#->';
preg_match_all("/(<-#)+([A-Za-z0-9_\s]+)(#->)+/s", $string, $matches);
$string = '<-#' . $matches[2][0] . '#->';
var_dump($string);
which outputs:
string(47) "<-#Any alphanumeric input that user may wish#->"
var_dump($matches); would return:
array(4) {
[0]=>
array(1) {
[0]=>
string(59) "<-#<-#<-#Any alphanumeric input that user may wish#->#->#->"
}
[1]=>
array(1) {
[0]=>
string(3) "<-#"
}
[2]=>
array(1) {
[0]=>
string(41) "Any alphanumeric input that user may wish"
}
[3]=>
array(1) {
[0]=>
string(3) "#->"
}
}