如何解析 XML 元素根和 select 它在 Java 中的某些值？

Question

我正在寻找一种实用的方法来解析 xml 根元素并从中获取一些值。我尝试了很多方法，但是 none 是有效的。

       DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();

                Document doc = factory.newDocumentBuilder().parse(fileLoaded);

                Element root = null;

                NodeList list = doc.getChildNodes();
               System.out.println(list.toString());
                for (int i = 0; i < list.getLength(); i++) {
                  if (list.item(i) instanceof Element) {
                    root = (Element) list.item(i);
                    System.out.println(root.toString());
                    break;
                  }
                }
                root = doc.getDocumentElement();
              }

XML 文件：

    <planes_for_sale id="planeId" num="1452" name="boing">
    <ad>
      <year> 1977 </year>
      <make> test </make>
      <model> Skyhawk </model>
      <color> Light blue and white </color>
      <description> New paint, nearly new interior,
        685 hours SMOH, full IFR King avionics </description>
      <price> 23,495 </price>
      <seller phone = "555-222-3333"> Skyway Aircraft </seller>
      <location>
      <city> Rapid City, </city>
      <state> South Dakota </state>
  </location>

在我的例子中，我想从 planes_for_sale 根元素加载 id、num、name。

Answer 1

尝试编组和解组机制

JAXBContext jaxbContext = JAXBContext.newInstance(EmployeeMap.class);
    Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
    EmployeeMap empMap = (EmployeeMap) jaxbUnmarshaller.unmarshal( new File("c:/temp/employees.xml") );

https://howtodoinjava.com/jaxb/jaxb-example-marshalling-and-unmarshalling-hashmap-in-java/

Answer 2

任何正在寻找实用方法的人都可以在这里进行测试：

    DocumentBuilderFactory factory = DocumentBuilderFactory
                    .newInstance();

            Document doc = factory.newDocumentBuilder().parse(fileLoaded);
            doc.getDocumentElement().normalize();
            System.out.println("Root element :"
                    + doc.getDocumentElement().getNodeName());
            NodeList nList = doc.getElementsByTagName("planes_for_sale");

              System.out.println("----------------------------");

                 for (int temp = 0; temp < nList.getLength(); temp++) {
                    Node nNode = nList.item(temp);
                    System.out.println("\nCurrent Element :" + nNode.getNodeName());
                    if (nNode.getNodeType() == Node.ELEMENT_NODE) {
                        Element eElement = (Element) nNode;
                        System.out.println("Plane name : " 
                           + eElement.getAttribute("name"));
                    }
                 }

Answer 3

使用 XPath 提取属性值和元素内容。

    DocumentBuilder builder = DocumentBuilderFactory
                                 .newInstance()
                                 .newDocumentBuilder();

    Document doc = builder.parse(...);

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

    String id = xp.evaluate("planes_for_sale/@id", doc);
    String num = xp.evaluate("planes_for_sale/@num", doc);
    String name = xp.evaluate("planes_for_sale/@name", doc);

    System.out.println("id: " + id);
    System.out.println("num: " + num);
    System.out.println("name: " + name);

生产：

id: planeId
num: 1452
name: boing

Answer 4

看看这个回答 How to read XML using XPath in Java

它告诉你如何

将 XML 文件读入 DOM
用XPath筛选出一组节点
对每个提取的节点执行特定操作。

可在此处找到支持 XPath 2.0 的变体

如何解析 XML 元素根和 select 它在 Java 中的某些值？

How to parse an XML element root an select some values of it in Java?

java

xml

jdom