的FtpWebRequest上的文件名

Question

的服務器和返回列表讀取文件我創建了一個方法，使閱讀從FTP服務器上的.xml的*擴展名的所有文件名，並返回字符串的文件名的數組。但我得到了一些錯誤「給出的路徑的格式不支持。」這裏是代碼片段：

protected static string[] FTPRelativePaths(string[] filelist) 
{ 
    if (null == filelist) 
     return new string[0]; 

    string[] result = new string[filelist.Length]; 
    for (int index = 0; index < filelist.Length; ++index) 
     result[index] = Path.GetFileName(filelist[index]); 

    return result; 
} 

public static string[] GetFileList(Uri serverUri, string ftpUserID, 
    string ftpPassword) 
{ 
    // The serverUri parameter should start with the ftp:// scheme. 
    if (serverUri.Scheme != Uri.UriSchemeFtp) 
    { 
     return null; 
    } 

    string[] fileList= new string[0]; 
    try 
    { 
     WebRequest.DefaultWebProxy = null; 
     FtpWebRequest ftprequest =(FtpWebRequest)WebRequest.Create(serverUri); 
     //ftprequest.UseBinary = true; 
     ftprequest.Method = WebRequestMethods.Ftp.ListDirectory; 
     ftprequest.Credentials =new NetworkCredential(ftpUserID, ftpPassword); 
     FtpWebResponse response = ftprequest.GetResponse() as FtpWebResponse; 
     string ext = "*.xml"; 
     string[] foundfiles = FTPRelativePaths(
      Directory.GetFiles(serverUri.ToString(), ext)); 

     if (foundfiles.Length > 0) 
     { 
      string[] newlist = new string[foundfiles.Length]; 
      foundfiles.CopyTo(newlist, fileList.Length); 
      fileList = newlist; 
      System.Array.Sort(fileList); 
     } 

     response.Close(); 
     return fileList; 
    } 
    catch (Exception ex) 
    { 
     System.Windows.Forms.MessageBox.Show(ex.Message); 
     fileList = null; 
     return fileList; 
    } 
} 

一旦我得到陣列中的所有文件名，我會結合成組合框sitesXmlCombo對象這一點，像這樣：

Uri uri = new Uri(uriString);    
string[] xmlFiles = FileList.GetFileList(uri, "username", "password"); 

sitesXmlCombo.DataSource = xmlFiles; 
Cursor.Current = Cursors.Arrow; 

我該如何做到這一點？我感謝您的幫助...

Answer 1

我覺得你有開放的錯誤;試試這個：

string _remoteHost = "ftp://ftp.name.com/htdocs/directory/"; 
FtpWebRequest request = (FtpWebRequest)WebRequest.Create(new Uri(_remoteHost));