Question

c#解析XML字符串

字符串内容如下："<?xml version=\"1.0\" encoding=\"GB2312\"?><InHosptMainTell xslFileName=\"InHosptMainTell_20040210095107.xsl\"><MainTell>胸闷3天，加重伴晕厥3小时。</MainTell></InHosptMainTell>" 取<MainTell>节点的内容（胸闷3天，加重伴晕厥3小时。）
xml是保存在ORACLE数据库中 C#解析后返回DataSet
（DataSet要包含XML各节点的名称和节点值）
<?xml version="1.0" encoding="GB2312"?><MHisDescript><individualMHis><txt_MRHistory>否认药物及食物过敏史。</txt_MRHistory></individualMHis><FaultHis_blood><txt_MRHistory>患者诉2年前开始出现双眼视物模糊。</txt_MRHistory></elapseMHis><FamilyHistory><txt_MRHistory>生于并长居广州，否认疫水接触史。</txt_MRHistory></FamilyHistory><MensesHistory><txt_MRHistory>平素生活起居有规律</txt_MRHistory></MensesHistory></MHisDescript>

Answer 1

private string SickNess(string XmlFile)
{
DataSet ds = new DataSet();
ds.ReadXml("sick.xml");
return ds.Tables[0].Rows[0][0].ToString();
}
用法：先把你的xml修改一下，去掉其中的转义字符\。存为sick.xml（如果用浏览器能打开，说明格式正确了。）然后上面的函数返回的就是你想要的字符串。

Answer 2

System.Xml.XmlDocument xd = new System.Xml.XmlDocument();
xd.LoadXml("<?xml version=\"1.0\" encoding=\"GB2312\"?><InHosptMainTell xslFileName=\"InHosptMainTell_20040210095107.xsl\"><MainTell>胸闷3天，加重伴晕厥3小时。</MainTell></InHosptMainTell>");
Response.Write(xd.SelectSingleNode("InHosptMainTell/MainTell/text()").Value);

Answer 3

string filePath = "e:\\11.xml";
XmlDocument xmlDoc = new XmlDocument();
xmlDoc.Load(filePath);
XmlNode root=xmlDoc.SelectSingleNode("InHosptMainTell");
XmlNode node = root.SelectSingleNode("MainTell");

Answer 4

看看 linq to xml 操作简单。