从Output生成xml文件

Question

到目前为止，我编写了一个代码，用于从ftp服务器下载文件，然后使用第三方dll（媒体信息）获取有关该文件的元数据详细信息。直到这很好，现在我正在尝试根据我的输出生成xml文件，我在这里看到了很好的例子How can I build XML in C#?来生成xml，但我的场景有点不同，这就是为什么我创建了这个线程。

下面是获取jpg文件属性值的代码

 static void Main(string[] args)
 {
     try
     {

         string file = "test.jpg";
         FtpWebRequest reqFTP;
         string ftpserverIp = "1.0.0.1";
         string fileName = @"c:\downloadDir\" + file;
         FileInfo downloadFile = new FileInfo(fileName);
         FileStream outputStream = new FileStream(fileName, FileMode.Append);
         reqFTP = (FtpWebRequest)FtpWebRequest.Create(new Uri("ftp://" + ftpserverIp + "/" + file));
         reqFTP.Method = WebRequestMethods.Ftp.DownloadFile;
         reqFTP.UseBinary = true;
         reqFTP.KeepAlive = false;
         reqFTP.Timeout = -1;
         reqFTP.UsePassive = true;
         reqFTP.Credentials = new NetworkCredential("sh", "SE");
         FtpWebResponse response = (FtpWebResponse)reqFTP.GetResponse();
         Stream ftpStream = response.GetResponseStream();
         long cl = response.ContentLength;
         // reqFTP.Method = WebRequestMethods.Ftp.ListDirectory;
         int bufferSize = 4096;
         int readCount;
         byte[] buffer = new byte[bufferSize];
         readCount = ftpStream.Read(buffer, 0, bufferSize);
         Console.WriteLine("Connected: Downloading File");
         while (readCount > 0)
         {
             outputStream.Write(buffer, 0, readCount);
             readCount = ftpStream.Read(buffer, 0, bufferSize);
             Console.WriteLine(readCount.ToString());
         }

         ftpStream.Close();
         outputStream.Close();
         response.Close();
         Console.WriteLine("Downloading Complete");
         var message = new StringBuilder();
         ConsoleApplication2.Program TechMeta = new ConsoleApplication2.Program();
         TechMeta.PutMessage(file, message);
         Console.WriteLine(message);
     }
     catch (Exception ex)
     {
         Console.Write(ex);
     }


 }

 private void PutMessage(string filename, StringBuilder message)
 {

     //Check the file is video or Image file here
     string extension =".jpg";

     bool b;
     if (b = filename.Contains(extension))
     {

         HowToUse_Dll.ImageInterrogator imageInterrogator = new HowToUse_Dll.ImageInterrogator();
         imageInterrogator.LoadFile(filename);
         message.AppendFormat(messageFormat, "Width", imageInterrogator.GetWidth(), Environment.NewLine);
         message.AppendFormat(messageFormat, "Height", imageInterrogator.GetHeight(), Environment.NewLine);
         message.AppendFormat(messageFormat, "FileSize", imageInterrogator.GetFileSize(), Environment.NewLine);
         message.AppendFormat(messageFormat, "FileFormat", imageInterrogator.GetFileFormat(), Environment.NewLine);
         message.AppendFormat(messageFormat, "Resolution", imageInterrogator.GetResolution(), Environment.NewLine);
     }
     else
     {

          HowToUse_Dll.VideoInterrogator videoInterrogator = new HowToUse_Dll.VideoInterrogator();
          videoInterrogator.LoadFile(filename);
          message.AppendFormat(messageFormat, "FileSize", videoInterrogator.GetFileSize(), Environment.NewLine);
          message.AppendFormat(messageFormat, "Duration", videoInterrogator.GetDuration(), Environment.NewLine);
          message.AppendFormat(messageFormat, "AspectRatio", videoInterrogator.GetAspectRatio(), Environment.NewLine);
          message.AppendFormat(messageFormat, "GetAspectRatio", videoInterrogator.GetAspectRatio(), Environment.NewLine);
          message.AppendFormat(messageFormat, "BitRate", videoInterrogator.GetBitRate(), Environment.NewLine);
          message.AppendFormat(messageFormat, "Format", videoInterrogator.GetFormat(), Environment.NewLine);
          message.AppendFormat(messageFormat, "VideoCoder", videoInterrogator.GetVideoCoder(), Environment.NewLine);
          message.AppendFormat(messageFormat, "Redirector", videoInterrogator.GetRedirector(), Environment.NewLine);
          message.AppendFormat(messageFormat, "TargetPlayback", videoInterrogator.GetTargetPlayback(), Environment.NewLine);
     }

 }

 public string messageFormat
 {
     get
     {
         return "{0}: {1}{2}";
     }
 }

基于test.jpg文件，TechMeta.PutMessage（文件，消息）; 消息值为

    {Width: 1024
     Height: 576
     FileSize: 84845
     FileFormat: JPEG
     Resolution: 8
     }

现在我正在尝试根据下面的值生成xml文件，更重要的是我尝试根据下面的新文件追加

<image Name="test.jpg">
<width>1024</width>
<height>576</height>
<file-Size>84845</file-Size>
<resolution>8</resolution>
<image Name="test1.jpg">
<width>1024</width>
<height>576</height>
<file-Size>84845</file-Size>
<resolution>8</resolution>

请提出任何建议

Answer 1

您可能需要查看课程XmlWriter。

编辑：错误的链接，现在修复。