Parsing a HTML file for CSS references

Question

I have a script which is suppose to collect all the css form a defined url or page. I have tried everything and for some reason it will not get it to detect linked stylesheets such as

<link rel="stylesheet" href="css/typography.css" />

I have tried everything I can think of. This is the code I am using that collects on page css and imports. Any help adding the link system would be great.

function scan($page_content){
    $i = 0;
    if(ereg("<style( *[\n]*.*)>\n*(.\n*)*<\/style>", $page_content)){
        if(preg_match_all("/(@\s*import\s* (url((\"|')?)?((\"|')?)|(\"|'){1}).+(\"|')?\)?)/", $page_content, $ext_stylesheets)){
            foreach($ext_stylesheets[0] as $stylesheet){
                $css_content[$i] = preg_replace("/(@\s*import\s*)|(url\(?((\"|')?))|(\"|'){1}|\)?(\"|')?;|(\s)/", "", $stylesheet);
                $i++;
            }
            $array = 1;
        }
        $inline_notused = $this->check_file($page_content, $page_content);
    }
    else die("No page styles, sorry!".$this->helptext);
}

I believe you could use xPath to parse the DOM rather than regex? — Luca
– Luca, Commented Oct 11, 2012 at 19:50
And if you must use regex, don't use ereg. it's deprecated and will be removed from PHP at some point. — Marc B
– Marc B, Commented Oct 12, 2012 at 20:49
What do you want the result to be with linked stylesheets? Do you want them to be @imported, or do you want the script to get the contents of the style? — uınbɐɥs
– uınbɐɥs, Commented Oct 14, 2012 at 5:50

uınbɐɥs · Accepted Answer · 2012-10-14 05:58:46Z

1

Here's a nice DOM/XPath way (demo):

function scan($html) {
    $dom = new DOMDocument;
    $dom->loadHTML($html);
    $path = new DOMXPath($dom);
    $nodes = $path->query('//style|//link');
    $style = '';
    foreach($nodes as $node) {
        if($node->tagName === 'style') {
            $style .= $node->nodeValue;
        } elseif($node->tagName === 'link') {
            $style .= "@import url('{$node->getAttribute('href')}')";
        } else {
            // Invalid
        }
        $style .= PHP_EOL;
    }
    return $style;
}

answered Oct 14, 2012 at 5:58

uınbɐɥs

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2 Comments

cHao Over a year ago

Kinda looks like you need to parse the <style> elements for imports and remove them, rather than including the whole content.

uınbɐɥs Over a year ago

@cHao Ah, I just looked at the code more closely, and it appears that he wants to get every stylesheet URL.

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