Question

I am trying to read a file name current.conf and then use the name of a folder saved in it to opendir(); when I open:

$file = fopen("current.conf","r");
$lines = fread($file,"10");
fclose($file);

$lines = "/".$lines."/";

echo $lines;

$dir=opendir($lines);

$files=array();
while (($file=readdir($dir)) !== false)
{
if ($file != "." and $file != ".." and $file != "index.php")
{
array_push($files, $file);
}
}
closedir($dir);

The current.conf has only one line in it:

2.1-2328

I am not able to open the folder that is named in the conf files. I have a feeling it has to do with the formatting of the conf file but not sure.

Answer 1

I suspect the directory doesn’t exist (or you don’t have the rights to read it), but without a specific error (opendir is most likely throwing an E_WARNING – check your logs, etc.)

Incidentally, you could re-write your code to reduce its complexity as follows:

<?php
    // Grab the contents of the "current.conf" file, removing any linebreaks.
    $dirPath = '/'.trim(file_get_contents('current.conf')).'/';

    $fileList = scandir($dirPath);

    if(is_array($fileList)) {
        foreach($fileList as $file) {
            // Skip the '.' and '..' in here as required.
            echo $file."\n";
        } 
    }
    else echo $dirPath.' cound not be scanned.';
?>

In this instance the call to scandir will throw an E_WARNING.