用java解析xml

Question

我是 xml 解析新手，我正在尝试使用 java 解析以下 xml 文件。

<a>
<e class="object">
    <amenities class="array">
        <e class="object">
            <id type="number">31</id>
            <name type="string">Internet access available</name>
        </e>
        <e class="object">
            <id type="number">9</id>
            <name type="string">Business center</name>
        </e>

</amenities>
<brands class="array">
        <e class="object">
            <code type="number">291</code>
            <name type="string">Utell</name>
        </e>
        <e class="object">
            <code type="number">72</code>
            <name type="string">Best Western International</name>
        </e>


</brands>
<hotels class="array">
        <e class="object">
            <addressLine1 type="string">4 Rue du Mont-Thabor</addressLine1>             
            <city type="string">Paris</city>                
            <name type="string">Renaissance Paris Vendome Hotel</name>
            <starRating type="string">5</starRating>                
        </e>
        <e class="object">
            <addressLine1 type="string">35 Rue de Berri</addressLine1>              
            <city type="string">Paris</city>                
            <name type="string">Crowne Plaza Hotel PARIS-CHAMPS ELYSÉES</name>
            <starRating type="string">5</starRating>                
        </e>
</hotels>    

</e>
</a>

我只需要列出我使用以下代码的名称标签信息（这将是酒店名称），但它不仅给我提供了酒店信息，还提供了所有内容，任何人都可以帮我解析这个吗？？？

多谢！！！

这是我使用的java代码

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.w3c.dom.Node;
import org.w3c.dom.Element;
import java.io.File;


public class ReadXMLFile {

public static void main(String argv[]) { 

 try {

    File fXmlFile = new File("c:\\file.xml");
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
    Document doc = dBuilder.parse(fXmlFile);
    doc.getDocumentElement().normalize();

    System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
    NodeList nList = doc.getElementsByTagName("e");
    System.out.println("-----------------------");

    for (int temp = 0; temp < nList.getLength(); temp++) {

       Node nNode = nList.item(temp);       
       if (nNode.getNodeType() == Node.ELEMENT_NODE) {

          Element eElement = (Element) nNode;

          System.out.println("Hotel Name : "  + getTagValue("name",eElement));


        }
    }
  } catch (Exception e) {
    e.printStackTrace();
  }
 }

 private static String getTagValue(String sTag, Element eElement){
    NodeList nlList= eElement.getElementsByTagName(sTag).item(0).getChildNodes();
    Node nValue = (Node) nlList.item(0); 

    return nValue.getNodeValue();    
 }

}

Answer 1

您可以使用 xpath 获取所有 name 节点：

XPath xpath = XPathFactory.newInstance().newXPath();
XPathExpression expr = xpath.compile("/hotels//name");
Object result = expr.evaluate(doc, XPathConstants.NODESET);
NodeList nodes = (NodeList) result;

Answer 2

Pavithira 您可以使用 xpath 仅获取酒店，下面是主要方法，您可以简单地复制/粘贴到您的代码中。

public static void main(String argv[]) {
        try {

               File fXmlFile = new File("c:\\file.xml");

        DocumentBuilderFactory dbFactory = DocumentBuilderFactory
                .newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(fXmlFile);
        doc.getDocumentElement().normalize();

        System.out.println("Root element :"
                + doc.getDocumentElement().getNodeName());
        XPathFactory factory = XPathFactory.newInstance();
        XPath xpath = factory.newXPath();
        XPathExpression expr = xpath.compile("//hotels/e");
        NodeList nList = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
        System.out.println("-----------------------");
        for (int temp = 0; temp < nList.getLength(); temp++) {
            Node nNode = nList.item(temp);
            if (nNode.getNodeType() == Node.ELEMENT_NODE) {
                Element eElement = (Element) nNode;
                System.out.println("Hotel Name : "
                        + getTagValue("name", eElement));
            }
        }
    } catch (Exception e) {
        e.printStackTrace();

    }

Answer 3

您正在迭代“e”节点，因此循环将打印出任何“e”节点（包括（子）根节点！）内的每个节点。如果您只想检索和打印这些节点，请将 getElementsByTagName 参数更改为“name”。