在C#中使用XPath从XML获取所有节点值

Question

我有以下XML:

<tcm:Component ID="tcm:481-636667" IsEditable="false" xmlns:tcm="http://www.tridion.com/ContentManager/5.0" xmlns:xlink="http://www.w3.org/1999/xlink">
  <tcm:Context>
    <tcm:Publication xlink:type="simple" xlink:title="07 Internal Test Publication" xlink:href="tcm:0-481-1"/>
    <tcm:OrganizationalItem xlink:type="simple" xlink:title="System Resources" xlink:href="tcm:481-92640-2"/>
  </tcm:Context> 
  <tcm:Data>
    <tcm:Title>IBE - Skywards</tcm:Title>
    <tcm:Type>Normal</tcm:Type>
    <tcm:Schema xlink:type="simple" xlink:title="Resources" xlink:href="tcm:481-190471-8"/>
    <tcm:Content>
      <Resources xmlns="http://www.sdltridion.com/tridion/schemas">
        <Text>
          <Key>SKYRL_MBD</Key>
          <Value>Miles Breakdown</Value>
        </Text>
        <Text>
          <Key>ltSR_MB.Text</Key>
          <Value>View Miles Breakdown</Value>
        </Text>
        <Text>
          <Key>ltSR_HMB.Text</Key>
          <Value>Hide Miles Breakdown</Value>
        </Text>
        <Text>
          <Key>SKYRL_MBD_LK</Key>
          <Value>Miles Breakdown</Value>
        </Text>    
      </Resources>
    </tcm:Content>
    <tcm:Metadata>
      <Metadata xmlns="http://www.sdltridion.com/tridion/schemas">
        <Language>
          <Language>English</Language>
        </Language>
      </Metadata>
    </tcm:Metadata>  
  </tcm:Data>
</tcm:Component>

现在我想在C#中编写一个方法,它将此XML作为输入,并将返回List中的所有"Key"和"Value"数据.

请建议.

Answer 1

首先,声明列表以保持值:

using System.Collections.Generic;

List<string> keysList = new List<string>();
List<string> valuesList = new List<string>();

然后:

using System.Xml; // System.Xml.dll

XmlDocument doc = new XmlDocument();
doc.LoadXml(xml); // Load(file)

var ns = new XmlNamespaceManager(doc.NameTable);
ns.AddNamespace("tcm", "http://www.tridion.com/ContentManager/5.0");
ns.AddNamespace("xlink", "http://www.w3.org/1999/xlink");

foreach (XmlNode node in doc.SelectNodes("//*[local-name()=\"Key\"]"))
{
    keysList.Add(node.InnerText);
}
foreach (XmlNode node in doc.SelectNodes("//*[local-name()=\"Value\"]"))
{
    valuesList.Add(node.InnerText);
}   

如果您不需要XML DOM,则只需要XPath来评估:

using System.Xml.XPath; // System.Xml.dll

XPathDocument doc = null;
using (TextReader reader = new StringReader(xml))
{
    doc = new XPathDocument(reader); // specify just path to file if you have such one
}
XPathNavigator nav = doc.CreateNavigator();
foreach (XPathNavigator node in (XPathNodeIterator)nav.Evaluate("//*[local-name()=\"Key\"]"))
{
     keysList.Add(node.Value);
}
foreach (XPathNavigator node in (XPathNodeIterator)nav.Evaluate("//*[local-name()=\"Value\"]"))
{
     valuesList.Add(node.Value);
}