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”？所以我需要用{src网站从上面}替换PLACEHOLDER

我找到了代码片段，我可以替换src链接但不用替换的url替换它。

如何做到这一点？

Answer 1

假设您在变量中包含源代码，您可以这样做：

$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);

PHP替换src链接并保留url

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