生成XML文档时出错 [英] There was an error generating the XML document

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本文介绍了生成XML文档时出错的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

您好

我看到"生成XML文档时出错"当我运行我的程序而我无法理解为什么?我是C#的新手

I see "There was an error generating the XML document" when I run my program and I cant understand why?I am new in C#

请告知

public
class
SerializableData

{

   public
void
save(
string
filename)
字符串
tmpFileName;

   public void save(string filename) String tmpFileName;

   {

   {

     

     

     tmpFileName =
filename +
" .tmp" ;
FileInfo
tmpFileInfo =
new
FileInfo (tmpFileName);
if
(tmpFileInfo.Exists)
FileStream
stream =
new
FileStream (tmpFileName,
< span style ="font-family:Consolas; color:#2b91af; font-size:x-small"> FileMode .Create);
true );

     tmpFileName = filename +".tmp"; FileInfo tmpFileInfo = new FileInfo(tmpFileName); if (tmpFileInfo.Exists) FileStream stream = new FileStream(tmpFileName, FileMode.Create); true);

     ; 

    

     

     

       tmpFileInfo.Delete();

       tmpFileInfo.Delete();

    

    

     save(stream);

     save(stream);

     stream.Close();

     stream.Close();

     tmpFileInfo.CopyTo(filename,

     tmpFileInfo.CopyTo(filename,

     tmpFileInfo .Delete();

     tmpFileInfo.Delete();

}

公开
void
保存(
Stream
stream)

public void save(Stream stream)

{

XmlSerializer
serializer =
new
XmlSerializer typeof SerializableData ));

XmlSerializer serializer = new XmlSerializer(typeof(SerializableData));

serializer.Serialize(stream,

serializer.Serialize(stream,

 

}

}

谢谢

推荐答案

你可以把一个try catch块放到抛出并粘贴exception.ToString()的代码里面。上面的代码也是不完整的。你能复制粘贴代码吗?再次?我尝试了下面的代码,它序列化了SerializableData类。

You can put a try catch block arond the code which throws and paste exception.ToString() here. Also the code above is incomplete. Can you please copy paste the code again? I tried following code and it serialized the class SerializableData.

 public class SerializableData
  {
    public int MyData { get; set; }
    public static void Save()
    {
      XmlSerializer ser = new XmlSerializer(typeof(SerializableData));
      ser.Serialize(Console.Out, new SerializableData() { MyData = 12 });
    }
  }

谢谢,

Neha Garg

Neha Garg


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