Preg_replace Pattern Variable

Question

I have a list of names and I'm using preg_replace to make a certain part of those names bold. I was using str_replace but I needed a limit on it.

However I'm getting the error "Delimiter must not be alphanumeric or backslash". After some research I found out that it's happening because I'm missing slashes on my pattern variable.

However I've been trying and I can't get it right and I'm not finding any example like mine. Thank you for your help.

while($row = $result->fetch_array()) {
        $name = $row['name'];

    $array = explode(' ',trim($name));
    $array_length = count($array);

    for ($i=0; $i<$array_length; $i++ ) {
        $letters = substr($array[$i], 0, $q_length);
        if ($letters = $q) {
        $bold_name = '<strong>'.$letters.'</strong>';
        $final_name = preg_replace($letters, $bold_name, $array[$i], 1);
        $array[$i] = $final_name; } 
    }

possible duplicate of [Delimiter must not be alphanumeric or backslash and preg\_match](http://stackoverflow.com/questions/7660545/delimiter-must-not-be-alphanumeric-or-backslash-and-preg-match) — HamZa, Jan 02 '15 at 11:22

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

You need a delimiter on your regex. For example, using ~:

$final_name = preg_replace('~'.$letters.'~', $bold_name, $array[$i], 1);

You can read more at the documentation for PCRE delimiters and the documentation for preg_replace.

From the docs:

When using the PCRE functions, it is required that the pattern is enclosed by delimiters. A delimiter can be any non-alphanumeric, non-backslash, non-whitespace character.

Often used delimiters are forward slashes (/), hash signs (#) and tildes (~). The following are all examples of valid delimited patterns.

/foo bar/

#^[^0-9]$#

+php+

%[a-zA-Z0-9_-]%

edited Jun 20 '20 at 09:12

Community

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answered Dec 28 '14 at 19:37

elixenide

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Oops. Copied and edited the wrong line. See my edit. – elixenide Dec 28 '14 at 19:42
Thank you very much, it was very simple after all! – Penny Dec 28 '14 at 19:46
Glad to help! Please remember to select an answer! :) – elixenide Dec 28 '14 at 19:46

score 0 · Answer 2 · answered Dec 28 '14 at 19:47

you just need to add simple delimiter to first parameter of preg_replace.

Add forward slashes to your $letters (However other delimiters can be used also.)

$final_name = preg_replace('/'.$letters.'/', $bold_name, $array[$i], 1);

As you mentioned in question, answer is in itself that Delimiter must not be alphanumeric or backslash.

Preg_replace Pattern Variable

2 Answers2