怎么读取XML里的值

Question

<?xml version="1.0" encoding="GB2312"?> 
<UploadFileInfo FileID="tttt" FileKey="eeee" FileName="aaa.doc" FileSize="118784" FileUploadTime="1316586458" FileSender="hehe" FileReceivers="xixi" />要把filename的值读出来,怎么写,谢谢了.

Answer 1

XmlDocument xml=new XmlDocument();
xml.Load(@"c:\abc.xml");
XmlNode node=xml.SelectSingleNode(@"//UploadFileInfo");
Console.Write(node.Attributes["FileName"].Value);

Answer 2

xml->XmlDocument->DataSet->DataTable
这样做做看，以前的帖子
http://topic.csdn.net/u/20091120/10/868e1689-3b38-4323-87d5-1899f6287e77.html

Answer 3

这里FileName 是属于节点UploadFileInfo 的特征了。所以要用node.Attributes["FileName"].Value

Answer 4

XML访问技术

Answer 5

xml.Load(@"c:\abc.xml",Encoding.Default);
或者
xml.Load(@"c:\abc.xml",Encoding.GetEncoding("gb2312"));