列出远程主机上的文件失败

Question

I have a code:

scandir("ssh2.sftp://" . intval($sftpHandle) . $remoteDir);

Why the same code works for one server but doesn't for another? There is a files on both servers. I can manage them via Filezilla without problems. The first one just returns array('.') even if there is a lot of files, another one returns array('file1', 'file2', 'file3', etc)

Even if I cannot list a files using scandir(), command ssh2_scp_recv($sshHandle, $remoteDir/file1, $localDir/file1) works fine. Also ssh2_exec($sshHandle, "ls $remoteDir") works fine to me.

Please note that I'm using $ ftpHandle for scandir() but $ sshHandle for ssh2_* functions. Using $sshHandle for scandir() cause "Segmentation fault" error.

I know that I can workaround this by parsing ssh2_exec($sshHandle, "ls $remoteDir") output, but would prefer do it right way if possible.

My PHP version is 7.0.31

Answer 1

This should work probably

$connection = ssh2_connect($url);

// login
if (!ssh2_auth_password($connection, $username, $password)) throw new Exception('Unable to connect server.');

// Create SFTP resource
if (!$sftp = ssh2_sftp($connection)) throw new Exception('Unable to create connection.');

$localDir  = '/path/to/local/dir';
$remoteDir = '/path/to/remote/dir';

// download all the files or list all
$files    = scandir('ssh2.sftp://' . $sftp . $remoteDir);

if (!empty($files)) {
  foreach ($files as $file) {
    if ($file != '.' && $file != '..') {


       // here you can print files using $file

       // or download file by uncommenting below line

      //ssh2_scp_recv($connection, "$remoteDir/$file", "$localDir/$file");

    }
  }
}