PHP- Preg_replace 标签名称及其内容

Question

我有一个像这样的字符串 string(8234) "<style>.{ margin-bottom:10px; float:left; display: inline-block; width:56%; text-align:left; padding-right:20px; padding-left:20px; } . > p { display: table-cell; height: 150px; vertical-align: middle; }..................</style>................. 我想删除 <style> 标签及其所有内容。我试过了

$description = $product_info['description']; // the string with style tag
$text = preg_replace('/<\s*style.+?<\s*\/\s*style.*?>/si', ' ', $description );
echo $text;

但它显示相同的字符串而不删除 <style> 标记。我还尝试通过

仅替换 <style> 标签

$text    = str_replace("<style>", "", $description);

但这也根本不起作用。我一次又一次地看到相同的字符串。我也尝试过 DOMParser

 $doc =   new DOMDocument();
 $doc->loadHTML($description);
 $xpath = new DOMXPath($doc);
 foreach ($xpath->query('//style') as $node) {
    $node->parentNode->removeChild($node);
 }
 $text    =    $doc->saveHTML();

但在所有情况下，输出与 <style> 及其内容相同

Answer 1

您的正则表达式 does work, but might I suggest a simpler pattern，使用非贪婪量词：

<style>.*?<\/style>

PHP- Preg_replace 标签名称及其内容

PHP- Preg_replace tag name and its contents

css

php

regex

preg-replace

str-replace