如何获得XML中某个节点的值，见下面的XML文件，在线等立结贴！

Question

<?xml version="1.0" encoding="UTF-8" ?>
<SOAP-ENV:Envelope xmlns:SOAP-ENV="http://schemas.xmlsoap.org/soap/envelope/" xmlns:SOAP-ENC="http://schemas.xmlsoap.org/soap/encoding/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema" SOAP-ENV:encodingStyle="http://schemas.xmlsoap.org/soap/encoding/" xmlns="http://www.monternet.com/dsmp/schemas/"><SOAP-ENV:Header>
<TransactionID xmlns="http://www.monternet.com/dsmp/schemas/" xsi:type="xsd:string">00140301801050</TransactionID> 
</SOAP-ENV:Header><SOAP-ENV:Body>
<UnSubscribeServiceResp xmlns="http://www.monternet.com/dsmp/schemas/">
<Version>1.5.0</Version>
<MsgType>UnSubscribeServiceResp</MsgType>
<hRet>115</hRet>
</UnSubscribeServiceResp>
</SOAP-ENV:Body></SOAP-ENV:Envelope>以上是XML文件问题：如何取得<hRet>节点的值？

Answer 1

读的sample,自己看下.
XmlDocument xmldoc = new XmlDocument();
String tempstr1 = Path.Combine(Server.MapPath("."),"xml");
String xmlPath = Path.Combine(tempstr1,"news_subject.xml");
xmldoc.Load(xmlPath);
XmlNodeList NdList = xmldoc.SelectNodes("NewsSubject/subject");
for(int i=0;i<NdList.Count;i++)
{
String lm_id = NdList[i].SelectSingleNode("lm_id").InnerXml.Trim();
String lm_name = "";
     lm_name = NdList[i].SelectSingleNode("lm_name").InnerXml.Trim();
Response.Write(lm_name+"<br>"+lm_id+"<br>");
}

Answer 2

xml的结构,自己看这个例子改把.
<?xml version="1.0" encoding="gb2312"?>
<NewsSubject>
  <subject>
    <lm_id>1</lm_id>
    <lm_name>人物0</lm_name>
  </subject>
  <subject>
    <lm_id>2</lm_id>
    <lm_name>春运1</lm_name>
  </subject>
  <subject>
    <lm_id>3</lm_id>
    <lm_name>地方撒法。</lm_name>
  </subject>
</NewsSubject>

Answer 3

你这段XML文件不是well formed.

Answer 4

<?xml version="1.0" encoding="UTF-8" ?>
<bookstore>
  <book ISBN="2-3631-4">
    <title>Oberon's Legacy</title>
    <author>Corets, Eva</author>
    <price>5.95</price>
  </book>
  <store>
    <a>123</a>
    <b>456</b>
  </store>
</bookstore>
代码如下(取a节点的值)：
Dim xmlDoc As New XmlDocument
        xmlDoc.Load("C:\test.xml")        Dim nodeList As XmlNodeList = xmlDoc.SelectSingleNode("bookstore").ChildNodes  '获取根节点下的所有子节点        Dim xn As XmlNode
        For Each xn In nodeList
            Dim xe As XmlElement = CType(xn, XmlElement)            Dim nls As XmlNodeList = xe.ChildNodes '继续获取xe子节点下的子节点
            Dim xn1 As XmlNode
            For Each xn1 In nls
                Dim xe2 As XmlElement = CType(xn1, XmlElement)
                If xe2.Name = "a" Then
                    Me.TextBox1.Text = xe2.InnerText
                    Exit For
                End If
            Next
        Next

Answer 5

各位大侠再帮帮忙！拜谢！出错信息：未将对象引用设置到对象的实例。以下是代码：//该returnValue的值就是获得xml字符串（内容见楼顶）
returnValue = PostXmlAndGetResponse(postToUrl,xmlBody);

XmlDocument doc = new XmlDocument();
doc.LoadXml(returnValue);XmlNamespaceManager nsmgr = new XmlNamespaceManager(doc.NameTable);
nsmgr.AddNamespace("SOAP-ENV","http://schemas.xmlsoap.org/soap/envelope/");

XmlNode node = doc.SelectSingleNode("SOAP-ENV:Envelope/SOAP-ENV:Body/UnSubscribeServiceResp/hRet",nsmgr);
string s = node.InnerText;if (s == "0")
Response.Write("退订成功!");
else
Response.Write("退订不成功:错误代码:" + s);哪里错了？在线等！等！等！