PHP: Get name of directories which doesn't contain file “input…”?

Question

Please give me a solution for get name of directories which doesn't contain file “input…” using php. My folder structure is like this:

Main Dir
  namedirectory.php
  Dir1
    File1
    input_1.txt
    File
  Dir2
    File5
    File6
  Dir3
    File1
    input_2.txt
  Dir4
    File1
    File2
  Dir5
    File1
    File2

Because Dir1 and Dir3 has "input.." files, so it will get name=dir2,dir4,dir5. I have code but it doesn't work. Please help me fix it !

<?php
$dir ='/directory';
while($dirs = glob($dir . '/*', GLOB_ONLYDIR)) {
  $dir .= '/*';
  if(!$d) {
     $d=$dirs;
   } else {
      $d=array_merge($d,$dirs);     
   }

  foreach ($d as $d) {

     if ($handle = opendir($d)) {
        while ($entry = readdir($handle)) {
           if (strpos($entry, "input") === FALSE) {
               echo $d;
               break;
           }
        }
      closedir($handle);
     }
  }

}
?>

Answer 1

Use tripple = because PHP will cast false to 0 which will be true.

Change:

if (strpos($entry, "input") == FALSE) {

Into

if (strpos($entry, "input") === FALSE) {

EDIT

$dir ='/directory';
while($dirs = glob($dir . '/*', GLOB_ONLYDIR)) {
  $dir .= '/*';
  if(!$d) {
     $d=$dirs;
   } else {
      $d=array_merge($d,$dirs);     
   }

  foreach ($d as $d) {

     if ($handle = opendir($d)) {
        $showDirectory = true;
        while ($entry = readdir($handle)) {
           if (strpos($entry, "input") !== FALSE) {
                $showDirectory = false;
                break;
           }
        }
        if($showDirectory)
        {
            echo $d;
        }
      closedir($handle);
     }
  }

}