从Output生成xml文件

So far i wrote a code to download a file from ftp server then using 3rd party dll (media info) to get the metadata details about the file. Until this am good,now am trying to generate the xml file based on my output ,i seen great examples here How can I build XML in C#? to generate xml but my scenario is little bit different,thats why i created this thread.

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，我在这里看到很好的例子如何在C＃中构建XML？生成xml，但我的场景有点不同，这就是我创建这个线程的原因。

below is the code to get the properties value of jpg file

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}";
     }
 }

base on the test.jpg file it ,TechMeta.PutMessage(file, message); Message value are

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

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

Now am trying to generate xml file based on the value like below,the more importantly am trying to append based on new file like below

现在我试图根据下面的值生成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>

Any suggestion please

有任何建议请

1 个解决方案

#1

You might want to take a look into the class XmlWriter.

您可能想要查看类XmlWriter。

Edit: Wrong Link, fixed now.

编辑：错误的链接，现在修复。