Conditionally Replace a specific Character in a string - php

Question

I am trying to remove the # sign from a block of text. The problem is that in certain cases (when at the beginning of a line, the # sign needs to stay.
I have succeeded by using the RegEx pattern .\#, however on when the # sign does get removed it also removes the character preceding it.
Goal: remove all # signs UNLESS the # sign is the first character in the line.
<?php
function cleanFile($text)
{
$pattern = '/.\#/';
$replacement = '%40';
$val = preg_replace($pattern, $replacement, $text);
$text = $val;
return $text;
};
$text = ' Test: test#test.com'."\n";
$text .= '#Test: Leave the leading at sign alone'."\n";
$text .= '#Test: test#test.com'."\n";
$valResult = cleanFile($text);
echo $valResult;
?>
Output:
Test: tes%40test.com
#Test: Leave the leading at sign alone
#Test: tes%40test.com

Answer 1

You can do this with regex using a negative lookbehind: /(?<!^)#/m (an # sign not preceded by the start of a line (or the start of the string if you skip out the m modifier)).
Regex 101 Demo
In code:
<?php
$string = "Test: test#test.com\n#Test: Leave the leading at sign alone\n#Test: test#test.com;";
$string = preg_replace("/(?<!^)#/m", "%40", $string);
var_dump($string);
?>
which outputs the following:
string(84) "Test: test%40test.com
#Test: Leave the leading at sign alone
#Test: test%40test.com;"
Codepad demo

Answer 2

There's no need for regexp in such simple case.
function clean($source) {
$prefix = '';
$offset = 0;
if( $source[0] == '#' ) {
$prefix = '#';
$offset = 1;
}
return $prefix . str_replace('#', '', substr( $source, $offset ));
}
and test case
$test = array( '#foo#bar', 'foo#bar' );
foreach( $test as $src ) {
echo $src . ' => ' . clean($src) . "\n";
}
would give:
#foo#bar => #foobar
foo#bar => foobar

Answer 3

the syntax [^] means negative match (as in don't match), but I don't think the following would work
$pattern = '/[^]^#/';