ftp url的opendir（）失敗

Question

我試圖從遠端服務器打開一個文件夾。我寫道：

if ($folderHandle = opendir($folder)) 

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

我得到奇怪的錯誤Warning: opendir(ftp://...:21): failed to open dir: operation failed in ... on line 38

任何想法，我應該何去何從？這是FTP憑證的問題嗎？

Answer 1

你可以使用PHP的FTP Capabilities遠程連接到服務器，並獲得一個目錄列表：

// 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);