使用NodeList遍历XML中的所有元素

Question

我想遍历XML打印中的所有元素。我的问题是，在staff1标记之后，我一直收到空指针异常，即john 465456433 gmail1 area1 city1

这是我的Java代码，用于打印xml文件中的所有元素：

File fXmlFile = new File("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("*");

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

Node n=null;
Element eElement=null;

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


  if (n.getNodeType() == Node.ELEMENT_NODE) {
    eElement = (Element) n.getChildNodes();
    System.out.println("\nCurrent Element :" + n.getNodeName());
    name = eElement.getElementsByTagName("name").item(i).getTextContent(); //here throws null pointer exception after printing staff1 tag
    phone = eElement.getElementsByTagName("phone").item(i).getTextContent();
    email = eElement.getElementsByTagName("email").item(i).getTextContent();
    area = eElement.getElementsByTagName("area").item(i).getTextContent();
    city = eElement.getElementsByTagName("city").item(i).getTextContent();
  }
  n.getNextSibling();
}

XML档案：

<?xml version="1.0"?>
<company>
  <staff1>
    <name>john</name>
    <phone>465456433</phone>
    <email>gmail1</email>
    <area>area1</area>
    <city>city1</city>
  </staff1>
  <staff2>
    <name>mary</name>
    <phone>4655556433</phone>
    <email>gmail2</email>
    <area>area2</area>
    <city>city2</city>
  </staff2>
  <staff3>
    <name>furvi</name>
    <phone>4655433</phone>
    <email>gmail3</email>
    <area>area3</area>
    <city>city3</city>
  </staff3>
</company>

预期产量：

john
465456433
gmail1
area1
city1
mary
4655556433
gmail2
area2
city2
furvi
4655433
gmail3
area3
city3

Answer 1

    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    DocumentBuilder db = dbf.newDocumentBuilder();
    Document dom = db.parse("file.xml");
    Element docEle = dom.getDocumentElement();
    NodeList nl = docEle.getChildNodes();
    int length = nl.getLength();
    for (int i = 0; i < length; i++) {
        if (nl.item(i).getNodeType() == Node.ELEMENT_NODE) {
            Element el = (Element) nl.item(i);
            if (el.getNodeName().contains("staff")) {
                String name = el.getElementsByTagName("name").item(0).getTextContent();
                String phone = el.getElementsByTagName("phone").item(0).getTextContent();
                String email = el.getElementsByTagName("email").item(0).getTextContent();
                String area = el.getElementsByTagName("area").item(0).getTextContent();
                String city = el.getElementsByTagName("city").item(0).getTextContent();
            }
        }
    }

遍历所有子项，nl.item(i).getNodeType() == Node.ELEMENT_NODE并用于过滤出文本节点。如果XML中没有其他内容，则剩下的是人员节点。

对于资料下的每个节点（名称，电话，电子邮件，区域，城市）

 el.getElementsByTagName("name").item(0).getTextContent();

el.getElementsByTagName("name")会在“东西”下提取“名称”节点，为
.item(0)您提供第一个节点，并.getTextContent()在其中获取文本内容。

编辑： 由于我们有杰克逊，所以我将以其他方式进行。为对象定义一个pojo：

public class Staff {
    private String name;
    private String phone;
    private String email;
    private String area;
    private String city;
...getters setters
}

然后使用杰克逊：

    JsonNode root = new XmlMapper().readTree(xml.getBytes());
    ObjectMapper mapper = new ObjectMapper();
    root.forEach(node -> consume(node, mapper));



private void consume(JsonNode node, ObjectMapper mapper) {
    try {
        Staff staff = mapper.treeToValue(node, Staff.class);
        //TODO your job with staff
    } catch (JsonProcessingException e) {
        e.printStackTrace();
    }
}