java - 获取java xml中节点的节点的节点？

Question

我没有太多的 java/xml 经验，所以这可能是微不足道的。但给定这样的 xml 结构:

<Things>
    <object name="cat" id="0">
        <prop id="1" name="race">ShortHair</prop>
    </object>

    <object name="car" id="1">
        <prop id="1" name="Manufacturer">
            <manufacturer id="1" name="ford">
        </prop>
    </object>

    <object id="2" name="Window">
    </object>
</Things>

是否可以获取特定节点，例如:

getObject("cat").getRace();

或

getObject("car").getId(1).getManufacturer();

或

getObject(id="2").name;

我知道，这不是有效的代码，但我没有找到似乎执行类似操作的 xml 示例。

谢谢。

Answer 1

是的，我们可以在 Java 中获取 XML 元素，并且有不同的方法。从教程的角度考虑这个示例。

XML:

<class>
   <student rollno = "393">
      <firstname>dinkar</firstname>
      <lastname>kad</lastname>
      <nickname>dinkar</nickname>
      <marks>85</marks>
   </student>

   <student rollno = "493">
      <firstname>Vaneet</firstname>
      <lastname>Gupta</lastname>
      <nickname>vinni</nickname>
      <marks>95</marks>
   </student>

   <student rollno = "593">
      <firstname>jasvir</firstname>
      <lastname>singh</lastname>
      <nickname>jazz</nickname>
      <marks>90</marks>
   </student>
</class>

Java:

import java.io.File;
import java.io.IOException;

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

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.w3c.dom.Node;
import org.w3c.dom.Element;
import org.xml.sax.SAXException;

public class XPathParserDemo {

   public static void main(String[] args) {

      try {
         File inputFile = new File("input.txt");
         DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
         DocumentBuilder dBuilder;

         dBuilder = dbFactory.newDocumentBuilder();

         Document doc = dBuilder.parse(inputFile);
         doc.getDocumentElement().normalize();

         XPath xPath =  XPathFactory.newInstance().newXPath();

         String expression = "/class/student";          
         NodeList nodeList = (NodeList) xPath.compile(expression).evaluate(
            doc, XPathConstants.NODESET);

         for (int i = 0; i < nodeList.getLength(); i++) {
            Node nNode = nodeList.item(i);
            System.out.println("\nCurrent Element :" + nNode.getNodeName());

            if (nNode.getNodeType() == Node.ELEMENT_NODE) {
               Element eElement = (Element) nNode;
               System.out.println("Student roll no :" + eElement.getAttribute("rollno"));
               System.out.println("First Name : " 
                  + eElement
                  .getElementsByTagName("firstname")
                  .item(0)
                  .getTextContent());
               System.out.println("Last Name : " 
                  + eElement
                  .getElementsByTagName("lastname")
                  .item(0)
                  .getTextContent());
               System.out.println("Nick Name : " 
                  + eElement
                  .getElementsByTagName("nickname")
                  .item(0)
                  .getTextContent());
               System.out.println("Marks : " 
                  + eElement
                  .getElementsByTagName("marks")
                  .item(0)
                  .getTextContent());
            }
         }
      } catch (ParserConfigurationException e) {
         e.printStackTrace();
      } catch (SAXException e) {
         e.printStackTrace();
      } catch (IOException e) {
         e.printStackTrace();
      } catch (XPathExpressionException e) {
         e.printStackTrace();
      }
   }
}

输出:

Current Element :student
Student roll no : 393
First Name : dinkar
Last Name : kad
Nick Name : dinkar
Marks : 85

Current Element :student
Student roll no : 493
First Name : Vaneet
Last Name : Gupta
Nick Name : vinni
Marks : 95

Current Element :student
Student roll no : 593
First Name : jasvir
Last Name : singh
Nick Name : jazz
Marks : 90

正如我所说，还有其他方法。检查this链接以获取更多此类示例。 XPath非常丰富。您可以使用 XPath 访问任何元素/属性。如果您想将 XML 消息转换为另一种 XML，您还可以使用 XSLT。