我使用以下代码将页面中的第四个图像作为og image tag回显，但是它使用Simple HTML DOM Parser，它首先读取页面然后重新加载编辑版本，这反过来减慢了页面的加载速度。有没有办法做同样的事情（从页面回显img标签）而不使用“简单的HTML DOM Parser”来重新加载页面？编辑此脚本或使用新脚本（如regex，preg_match，getElementsByTagName等）。
我可以使用JavaScript来执行此操作，但它将加载客户端，并且需要在加载页面时加载服务器端。
目前的php代码：
<?php
include('simple_html_dom.php');
$link="";
$html = file_get_html('/var/www/vhosts/example.com/httpdocs' .
$_SERVER["REQUEST_URI"]);
$link=$html->find('img')[3]->src;
$ret = $html->find('meta[property="og:image"]');
$ret[0] = new stdClass();
//$ret1[0] = new stdClass();
$ret[0]->content = $link;
?>
<meta property="og:image"
content="https://example.com/<?php echo $ret[0]->content; ?>" />
html示例：
<body>
<p> </p>
<p> </p>
<p>Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod
tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,
quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo
consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum
dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident,
sunt in culpa qui officia deserunt mollit anim id est laborum. </p>
<p>
<img border="0" src="../path/to/file/image.jpg" width="1200" height="630"></p>
<p>Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod
tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,
<p>
<img border="0" src="../path/toanoter/file/imagetwo.jpg" width="1200" height="630"></p>
quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo
consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum
dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident,
sunt in culpa qui officia deserunt mollit anim id est laborum. </p>
<p>
<img border="0" src="../path/to/file/image345.jpg" width="1200" height="630"></p>
<p>Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do eiusmod
tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam,
quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo
<p>
<img border="0" src="../../path/imagenew.jpg" width="1200" height="630"></p>
consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum
dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident,
sunt in culpa qui officia deserunt mollit anim id est laborum. </p>
<p> </p>
</body>
输出应该是：
<meta property="og:image"
content="https://example.com/path/imagenew.jpg" />
你可以尝试这样的事情：
$pattern = '/src=\".+(\/path.+\.jpg)\"/m';
$html = 'your html page';
preg_match_all($pattern, $html, $matches, PREG_SET_ORDER, 0);
$siteUrl = 'https://example.com';
echo $siteUrl . $matches[3][1];