关于使用Java的XPath

Question

通过使用java提供XPath来获取元素的值时遇到问题。我尝试了很多东西，但都无法成功。

import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStream; 
import java.io.InputStreamReader; 
import java.io.StringReader; 
import java.util.HashMap; 
import java.util.Map; 

import javax.xml.parsers.ParserConfigurationException; 
import javax.xml.xpath.XPath; 
import javax.xml.xpath.XPathConstants; 
import javax.xml.xpath.XPathExpression; 
import javax.xml.xpath.XPathExpressionException; 
import javax.xml.xpath.XPathFactory; 

import org.w3c.dom.NodeList; 
import org.xml.sax.InputSource; 
import org.xml.sax.SAXException; 

import com.dell.logistics.framework.transform.NamespaceContext; 

public class GetXPath { 

    protected Object evaluate(String xpathStr, String xml, String namespaces) throws XPathExpressionException { 
     InputSource inputSource = new InputSource(new StringReader(xml)); 
     XPathFactory factory = XPathFactory.newInstance(); 
     XPath xPath = factory.newXPath(); 
     NamespaceContext nsContext = new NamespaceContext(); 
     nsContext.setNamespacesMap(getNsMap(namespaces)); 
      //System.out.println(nsContext.getPrefix(namespaces)); 
     xPath.setNamespaceContext(nsContext); 
     XPathExpression xpExp = xPath.compile(xpathStr); 
     return xpExp.evaluate(inputSource, XPathConstants.NODESET); 
    } 

    private Map<String, String> getNsMap(String namespaces) { 

     String delims = ","; 
     String[] nsKeyValue = namespaces.split(delims); 
     Map<String, String> mp = new HashMap<String, String>(); 
     for (String string : nsKeyValue) { 
      mp.put(string.split("=")[0], string.split("=")[1]); 
      System.out.println(string.split("=")[0] + string.split("=")[1]); 
     } 
     return mp; 
    } 

    public static String readFile(String fileName) { 
     try { 
     // InputStream is = null; 
       InputStream is = GetWorkOrderDataExtractor.class.getResourceAsStream(fileName); 
       BufferedReader br = new BufferedReader(new InputStreamReader(is)); 
       StringBuffer sb = new StringBuffer(); 
       String l = null; 
       while ((l = br.readLine()) != null) { 
        sb.append(l).append("\n"); 
       } 
       return sb.toString(); 
     } catch (Exception e) { 
       e.printStackTrace(); 
     } 
     return null; 
    } 


    public static void main(String[] args) 
    throws ParserConfigurationException, SAXException, 
      IOException, XPathExpressionException { 
     GetXPath g = new GetXPath(); 
     String xml = readFile("fooewo.xml"); 
     String value = null; 
     System.out.println(xml); 


     NodeList containerNodes = (NodeList) g.evaluate(
       "/demo",xml, 
       "a=http://schemas.demo.com/it/WorkOrderChannelAckNackResponse/1.0"); 

     try{ 
     for (int i = 0; i < containerNodes.getLength(); i++) { 

       // get the node value. 
       value = containerNodes.item(i).getTextContent(); 
       System.out.println(value); 
      } 
     System.out.println("Node Found : " + containerNodes.getLength() + " times"); 
     } 
     catch (Exception e) 
     { 
      e.printStackTrace(); 
     } 
    } 
} 
" 

XML文件：

<?xml version="1.0" encoding="utf-8"?> 
<demo xmlns="try with ur schema"> 
    <test> 
     <value>10</value> 
     <color>red</color> 
     <animal>dog</animal> 
     <day>13</day> 
     <age>22</age> 
    </test> 
    <test> 
     <value>20</value> 
     <color>green</color> 
     <animal>cat</animal> 
     <day>12</day> 
     <age>23</age> 
    </test> 
</demo> 

任何帮助表示赞赏。

感谢，
普拉迪普

Answer 1

我觉得很容易计算XPath的最好方法是使用AXIOMXPath

下面是一个例子，

OMElement documentElement = new StAXOMBuilder(inStreamToXML).getDocumentElement(); 
AXIOMXPath xpathExpression = new AXIOMXPath ("/demo"); 
List nodeList = (OMNode)xpathExpression.selectNodes(documentElement); 

通过遍历列表，你可以得到的结果容易。