php creating a css stylesheet list from directory

Question

Im trying to read css files from a directory and create the respective html markup, like

<link rel="stylesheet" type="text/css" media="screen" href="css/bootstrap.css">

PHP

 $css_dir = 'application/resources/css';

$fp = opendir($css_dir);
while ($file = readdir($fp)) 
{
if (strpos($file, '.css',1))
$results[] = $this->base.'/'.$file;                 
}
closedir($fp);

print_r($results);

this works correctly creating an array with results like [1] => Directory/name.css which is fine but if I change the results line to something like

$results[] = '<link rel="stylesheet" type="text/css" media="screen" href="'.$this->base.'/'.$file.'">';

I get am empty array, the problem is with the '<' '>' symbols is there no way to do this in one go without having to then do a foreach through the results array?

Be aware - and this is just a warning - that if you plan on making this a public-facing application, it is very insecure to blindly load files in a directory. Good luck — David
– David, Commented Dec 28, 2013 at 20:36
What? Did you check the source code, since <link ..> probably won't show on the screen - but nevertheless it should work. However, it can be simplified by using glob — kero
– kero, Commented Dec 28, 2013 at 20:37
@David can you explain how? A user will not be able to place any files into that directory without root access. — Edward
– Edward, Commented Dec 28, 2013 at 20:39
Try echo '<pre>' . htmlentities(print_r($results, true)) . '</pre>'; to see the results literally. — Barmar
– Barmar, Commented Dec 28, 2013 at 20:41
@Barmar thanks this is just what I needed, i forgot link will just get rendered and won't show up on the print_r. — Edward
– Edward, Commented Dec 28, 2013 at 20:42

Noor · Accepted Answer · 2013-12-28 20:56:25Z

1

Instead of using < and > use < and > accordingly, or htmlspecialchars()

Improvement in your functional code

 /* This is the WRONG way to loop over the directory. */
    while ($entry = readdir($handle)) {
        echo "$entry\n";
    }

 /* This is the correct way to loop over the directory. */
    while (false !== ($entry = readdir($handle))) {
        echo "$entry\n";
    }

Reference

edited Dec 28, 2013 at 20:56

answered Dec 28, 2013 at 20:43

Noor

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