ftp url的opendir()失败

Question

我正在尝试打开远程服务器上的文件夹。我写道：

if ($folderHandle = opendir($folder))

where $folder = "ftp://xxx:xxx@xxx.net:21"

我得到了奇怪的错误Warning: opendir(ftp://...:21): failed to open dir: operation failed in ... on line 38

我该怎么做，你有什么想法吗？FTP凭据有问题吗？

Answer 1

您可以使用PHP的FTP Capabilities远程连接到服务器并获取目录列表：

// set up basic connection
$conn_id = ftp_connect('otherserver.example.com'); 

// login with username and password
$login_result = ftp_login($conn_id, 'username', 'password'); 

// check connection
if ((!$conn_id) || (!$login_result)) { 
    echo "FTP connection has failed!";
    exit; 
}

// upload the file
$upload = ftp_put($conn_id, $destination_file, $source_file, FTP_BINARY); 

// check upload status
if (!$upload) { 
    echo "FTP upload has failed!";
} else {
    echo "Uploaded $source_file to $ftp_server as $destination_file";
}

// Retrieve directory listing
$files = ftp_nlist($conn_id, '/remote_dir');

// close the FTP stream 
ftp_close($conn_id);