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Home/ Questions/Q 8304715
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Editorial Team
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Editorial Team
Asked: June 8, 20262026-06-08T17:55:15+00:00 2026-06-08T17:55:15+00:00

I need help creating a regular expression blah blah blah blah <?= __l(array(‘key’=>’SOMEVALUE’,’default’=>’string string

  • 0

I need help creating a regular expression

blah blah blah blah <?= __l(array('key'=>'SOMEVALUE','default'=>'string string string')) ?>blah blah

I want to be able to strip the ‘SOMEVALUE’ and the ‘string string string’

Thanks
In advance

====== ‘This is what I have’ ========

$subject = "Blah blah ja ja blah blah jank junk jonk <?= __l(array('key'=>'KEYKEYKEY','default'=>'I am a monkey sigh\'s')) ?> ldjlsakfdj as;dfj as;flkj a fsd  ljaasfd <?= __l(array('key'=>'KEYKEYKEY','default'=>'I am a monkey sigh\'s')) ?>  ";
$pattern = '#\_\_l\(array\(\'key\'=>\'(.*)\',\'default\'=>\'(.*)\'\)\)#';



if (preg_match_all($pattern, $subject)) {
print "A match was found.";
} else {
print "A match was not found.";
}

print '<br />';

preg_match_all($pattern, $subject, $matches);

echo '<pre>';
print_r($matches);
echo '</pre>'
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  1. Editorial Team
    Editorial Team
    2026-06-08T17:55:16+00:00Added an answer on June 8, 2026 at 5:55 pm

    Your regex is a bit too complicated, try something like this:

    $regex = "/'key'=>'[^']+','default'=>'[^']+'/";
    $string = preg_replace( $regex, "'key'=>'','default'=>''", $string);
    

    The regex is:

    'key'=>'        - Match this literally
    [^']+           - Match anything that's not a single quote, one or more times
    ','default'=>'  - Match this literally
    [^']+           - Match anything that's not a single quote, one or more times
    '               - Match this literally
    

    And the replacement is simply the 2nd parameter to preg_replace().

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