Preg_replace and Preg_replace_callback

Question

I'm updating my coding and moving away from preg_replace towards the function preg_replace_callback. On the first I used two arrays, which replaced all the matches in a piece of text.

I have setup the following new piece of code, but I am running into some issues:

<?php
$inhoud = " dit is een test versie, waarin [alum] staat [alum] & [fotoalbums] om [intern=test]te[/intern] vervangen<p>";

function parse_callback($match) {
        $hit = $match[0];
        switch ($hit){
                case '[alum]': 
                        return "<a href=\"/linktype1/\">link1</a>";
                        break;
                case '[fotoalbums]':
                        return "<a href=\"/linktype2/\">link2</a>";
                        break;
                case '[intern]':
                        return "<a href=\"". $match[1] ."\">$match[2]</a>";
                        break;
                default:
                        //return "UNKNOWN:$match";
                        return var_dump($match);
        }
}

$Patroon = "'\[intern=(.*?)\](.*?)\\[\/intern\]'";
$Patroon = "'\[fotoalbums\]'";
$Patroon = "'\[alum\]'";

$inhoud = preg_replace_callback($Patroon, parse_callback, $inhoud);
?>

The later two in the $Patroon are no issue, these will be updated, but I'm unable to update the first, as the match being found will be the complete string. The $match will also contain the pieces of text from $inhoud on which is has matched (.*?), but I'm unable to process these in the switch.

Any suggestions on how to resolve this issue or a better approach to the coding. The above example is just a few options which will be matched, the actual list is much larger.

Mariano · Accepted Answer · 2015-11-08 10:24:50Z

2

First of all, you should pass the function name as a string: "parse_callback".

As you've noticed, match[0] will return the whole matched text ("[intern=test]te[/intern]"). You could capture the tag name in a group:

$Patroon = "'\[(intern)=(.*?)\](.*?)\[/intern\]'";

Notice the parentheses in (intern)

This way, the tag name will be returned by $match[1].

And using this approach to all patterns:

$Patroon = array(
    "'\[(intern)=(.*?)\](.*?)\[/intern\]'",
    "'\[(fotoalbums|alum)\]'"
);

Code

<?php
$inhoud = " dit is een test versie, waarin [alum] staat [alum] & [fotoalbums] om [intern=test]te[/intern] vervangen<p>";

function parse_callback($match) {
        //check the value of the first capture
        $hit = $match[1];
        switch ($hit){
                case 'alum': 
                        return "<a href=\"/linktype1/\">link1</a>";
                        break;
                case 'fotoalbums':
                        return "<a href=\"/linktype2/\">link2</a>";
                        break;
                case 'intern':
                        return "<a href=\"$match[2]\">$match[3]</a>";
                        break;
                default:
                        //return "UNKNOWN:$match";
                        return var_dump($match);
        }
}

$Patroon = array(
    "'\[(intern)=(.*?)\](.*?)\[/intern\]'",
    "'\[(fotoalbums|alum)\]'"
);

$inhoud = preg_replace_callback($Patroon, "parse_callback", $inhoud);

echo $inhoud;
?>

Output

 dit is een test versie, waarin <a href="/linktype1/">link1</a> staat <a href="/linktype1/">link1</a> & <a href="/linktype2/">link2</a> om <a href="test">te</a> vervangen<p>

ideone demo

edited Nov 8, 2015 at 10:24

answered Nov 6, 2015 at 7:59

Mariano

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3 Comments

Mariano Over a year ago

I'm glad. Also keep in mind you could easily merge the last 2 patterns into 1.

Ralf Over a year ago

How do you mean, you can merge the last two patterns into 1? Not sure if I understand completely what you mean.

Mariano Over a year ago

@Ralf They can both be merged as "'\[(fotoalbums|alum)\]'". I edited the answer to show what I meant.

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