PHP ftp_pwd returns empty - php

Question

I'm writing a php script that gets called via ajax in javascript. I'm tyring to connect to an ftp server and list all the files in the "public_html" folder.
I also wanted to print the current ftp dir but when when I try, it prints as empty in the console.
I can connect to the ftp server, but I can't tell what dir is the current one.
//Get ftp user/pass
$ftp_server = "ftp." . $_POST['hostname'];
$ftp_username = $_POST['user'];
$ftp_userpass = $_POST['pass'];
echo "console.log('attempting to connect to ftp host: $ftp_server, user: $ftp_username, password: $ftp_userpass');\n";
$ftp_conn = ftp_connect($ftp_server) or die("alert('Could not connect to $ftp_server');\n");
// turn passive mode on
//ftp_pasv($ftp_conn, true);
ftp_chdir($ftp_conn, 'public_html');
$login = ftp_login($ftp_conn, $ftp_username, $ftp_userpass);
// then do something...
echo "console.log('Connected to FTP');\n";
// get contents of the current directory
$curdir = ftp_pwd($ftp_conn);
echo "console.log('Current dir: $curdir');\n";
$contents = ftp_nlist($ftp_conn, ".");
// output $contents
echo "var files = [];";
foreach ($contents as $file){
echo "var tempFile = '$file';\n
files.push(tempFile);\n";
}
echo "console.log(files);\n";
// close connection
ftp_close($ftp_conn);
?>

Answer 1

swap these two lines from:
ftp_chdir($ftp_conn, 'public_html');
$login = ftp_login($ftp_conn, $ftp_username, $ftp_userpass);
to:
$login = ftp_login($ftp_conn, $ftp_username, $ftp_userpass);
ftp_chdir($ftp_conn, 'public_html');

Answer 2

I have some problems, when using ftp_pasv();
Try to comment this line:
ftp_pasv($ftp_conn, true);