CODEWITHSUNDEEP
phpregexpreg-replace
Afsar
Afsar

Reputation: 3124

Escaping words in Preg Match

I have a list of words (in array) to be highlighted in string ( $text ) .Here is my code

$text = " A string with a spans  and color highlighted";
$words = array('and','span');

    foreach($words as $word){
      $patterns[] = '/'.$word.'/i';
    }

    foreach($words as $word){
      $replacements[] = "<span style='color:red;font-weight:bold;'>".$word."</span>";
    }

echo preg_replace($patterns, $replacements, $text);

I want to replace words span and color from a $text, but the result is something different, it also replacing the html tag span. How can i overcome this issue . or can i have alternative to this.

You can reproduce problem here. http://writecodeonline.com/php/ enter image description here

Thanks in advance.

Upvotes: 0

Views: 65

Answers (2)

Cristofor
Cristofor

Reputation: 2097

Here is a simplier solution, using explode function:

$text = "A string with a span and color highlighted";
$words = array('and','span');
$exploded = explode(" ", $text);
$i = 0;
foreach ($exploded as $word) {
    if (in_array($word, $words)) {
        $exploded[$i] = "<span style='color:red;font-weight:bold;'>".$word."</span>";
    }
    $i++;
}
print_r($exploded);

RESULT:

    Array
(
    [0] => A
    [1] => string
    [2] => with
    [3] => a
    [4] => <span style='color:red;font-weight:bold;'>span</span>
    [5] => <span style='color:red;font-weight:bold;'>and</span>
    [6] => color
    [7] => highlighted
)

Upvotes: 0

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89547

You don't need to produce a pattern and a replacement string for each word of the list. You only need to build one pattern and one replacement string with a backreference:

$text = " A string with a span  and color highlighted";
$words = array('and', 'span');

$pattern = '~\b(?:' . implode('|', $words) . ')\b~';

$replacement  = '<span style="color:red;font-weight:bold;">$0</span>';

$result = preg_replace($pattern, $replacement, $text);

In the replacement string the backreference $0 refers to the whole match result.

Since you parse the string only once, you avoid the problem.

Upvotes: 1

Related Questions