PHP替换src链接并保留url

Question

我有这样的源代码：

<script type='text/javascript' src='http://html.com/wp-content/plugins/contact-form-7/includes/js/scripts.js?ver=4.6'></script>
<script type='text/javascript'></script>
<script type='text/javascript' src='http://html.com/wp-content/plugins/table-of-contents-plus/front.min.js?ver=1509'></script>
<script type='text/javascript' src='http://html.com/wp-includes/js/wp-embed.min.js?ver=4.7'></script>

如何将所有src属性替换为：

"http://www.example.com/site=PLACEHOLDER&somethingelse"？因此，我需要用{ src网站从上面}替换占位符。

我找到了可以替换src链接的代码片段，但没有将其替换为已替换的url本身。

怎么才能做到呢？

Answer 1

假设您在变量中有源代码，您可以：

$code = "
<script type='text/javascript' src='http://html.com/wp-content/plugins/contact-form-7/includes/js/scripts.js?ver=4.6'></script>
<script type='text/javascript'></script>
<script type='text/javascript' src='http://html.com/wp-content/plugins/table-of-contents-plus/front.min.js?ver=1509'></script>
<script type='text/javascript' src='http://html.com/wp-includes/js/wp-embed.min.js?ver=4.7'></script>";

$pattern = "/src='([^']+)/i";
$replacement = "src='http://www.example.com/site=$1&somethingelse";

echo preg_replace($pattern, $replacement, $code);