PHP cURL to return remote page stylesheets - php

Question

I'm using following code to get remote content using PHP cURL
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "http://example.com");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
$output = curl_exec($ch);
curl_close($ch);
echo $output;
This code returns whole content But I just want to print all stylsheets in following format.
<link rel="stylesheet" href="http://www.example.com/css/style1.css">
<link rel="stylesheet" href="http://www.example.com/css/style2.css">
How do I Filter content using str.replace() to only get stylsheets with cURL?

Answer 1

If you only want to leave the <link> elements intact then you can use PHP's strip_tags() function.
strip_tags — Strip HTML and PHP tags from a string
It accepts an additional parameter that defines allowed tags, so all you have to do is set the only allowed tag to be the <link> tag.
$output = curl_exec($ch);
$linksOnly = strip_tags($ouput,'link');
The main problem here is that you don't really know what content you are going to get and trying to parse HTML content with anything other than a tool designed for that task may leave you with grey hair and a nervious twitch ;)
References -
strip_tags()

Answer 2

Using simple html dom library,
include('simple_html_dom.php');
// get DOM from URL or file
$html = file_get_html('http://www.example.com/');
// or your can get $html string through your curl request and say
// $html = str_get_html($html);
// find all "link"
foreach($html->find('link') as $e) {
if($e->type="text/css" && strpos($e->href, ":/") !=== false) // you don't want relative css hrefs. right?
echo $e->href."<br>";
}

Answer 3

A better approach would be to use PHP DOM to parse the HTML tree and retrieve the required nodes - <link> in your case - and filter them appropriately.

Answer 4

Using a regex:
preg_match_all('/rel="stylesheet" href="(.*)">/', $output, $matches);
if (isset($matches[1]) && count($matches[1]))
{
foreach ($matches as $value)
{
echo '<link rel="stylesheet" href="'.$value.'">';
}
}