PHP DOM 获取网站所有脚本src

Question

我想使用 curl 和 DOM 从网站获取所有脚本 src 链接。

我有这个代码：

$scripts = $dom->getElementsByTagName('script');

foreach ($scripts as $scripts1) {

    if($scripts1->getAttribute('src')) {

        echo $scripts1->getAttribute('src');

    }

}

此脚本运行正常，但如果网站有这样的脚本标签会怎样：

<script type="text/javascript">
window._wpemojiSettings = {"source":{"concatemoji":"http:\/\/domain.com\/wp-includes\/js\/wp-emoji-release.min.js?ver=4.2.4"}}; ........
</script>

我还需要获取此脚本源。我该怎么做？

Answer 1

如果您的第一个解析器是空的，我会使用正则表达式创建另一个，即：

$html = file_get_contents("http://somesite.com/");

preg_match_all('/<script.*?(http.*?\.js(?:\?.*?)?)"/si', $html, $matches, PREG_PATTERN_ORDER);
for ($i = 0; $i < count($matches[1]); $i++) {
    echo str_replace("\/", "/", $matches[1][$i]);
}

您可能需要调整正则表达式以适用于不同的网站，但上面的代码应该让您了解您需要什么。

演示： http://ideone.com/Fwf6Mb

正则表达式解释：

<script.*?(http.*?\.js(?:\?.*?)?)"
----------------------------------

Match the character string “<script” literally «<script»
Match any single character «.*?»
   Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the regex below and capture its match into backreference number 1 «(http.*?\.js(?:\?.*?)?)»
   Match the character string “http” literally «http»
   Match any single character «.*?»
      Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
   Match the character “.” literally «\.»
   Match the character string “js” literally «js»
   Match the regular expression below «(?:\?.*?)?»
      Between zero and one times, as many times as possible, giving back as needed (greedy) «?»
      Match the character “?” literally «\?»
      Match any single character «.*?»
         Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the character “"” literally «"»

正则表达式教程

http://www.regular-expressions.info/tutorial.html

PHP DOM 获取网站所有脚本src

PHP DOM Get website all scripts src

javascript

php

dom

preg-match