如何在PHP中获取目录大小

Question

function foldersize($path) {
$total_size = 0;
$files = scandir($path);

  foreach($files as $t) {
    if (is_dir(rtrim($path, '/') . '/' . $t)) {
      if ($t<>"." && $t<>"..") {
          $size = foldersize(rtrim($path, '/') . '/' . $t);

          $total_size += $size;
      }
    } else {
      $size = filesize(rtrim($path, '/') . '/' . $t);
      $total_size += $size;
    }
  }
  return $total_size;
}

function format_size($size) {
  $mod = 1024;
  $units = explode(' ','B KB MB GB TB PB');
  for ($i = 0; $size > $mod; $i++) {
    $size /= $mod;
  }

  return round($size, 2) . ' ' . $units[$i];
}

$SIZE_LIMIT = 5368709120; // 5 GB

$sql="select * from users order by id";
$result=mysql_query($sql);

while($row=mysql_fetch_array($result)) {
  $disk_used = foldersize("C:/xampp/htdocs/freehosting/".$row['name']);

  $disk_remaining = $SIZE_LIMIT - $disk_used;
  print 'Name: ' . $row['name'] . '<br>';

  print 'diskspace used: ' . format_size($disk_used) . '<br>';
  print 'diskspace left: ' . format_size($disk_remaining) . '<br><hr>';
}

在脚本执行完成之前，为什么处理器使用率会过高或100％升高？可以做些什么来优化它吗？还是有其他方法可以检查文件夹和其中的文件夹大小？

Answer 1

以下是其他地方提供的其他解决方案：

如果在Windows主机上：

<?
    $f = 'f:/www/docs';
    $obj = new COM ( 'scripting.filesystemobject' );
    if ( is_object ( $obj ) )
    {
        $ref = $obj->getfolder ( $f );
        echo 'Directory: ' . $f . ' => Size: ' . $ref->size;
        $obj = null;
    }
    else
    {
        echo 'can not create object';
    }
?>

否则，如果在Linux主机上：

<?
    $f = './path/directory';
    $io = popen ( '/usr/bin/du -sk ' . $f, 'r' );
    $size = fgets ( $io, 4096);
    $size = substr ( $size, 0, strpos ( $size, "\t" ) );
    pclose ( $io );
    echo 'Directory: ' . $f . ' => Size: ' . $size;
?>