我想循环遍历每个XML打印的所有元素。我的问题是我在staff1
标记之后不断获得空指针异常,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文件:
<?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>
预期产出:
john
465456433
gmail1
area1
city1
mary
4655556433
gmail2
area2
city2
furvi
4655433
gmail3
area3
city3
答案 0 :(得分:64)
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();
}
}
}
迭代所有孩子,nl.item(i).getNodeType() == Node.ELEMENT_NODE
用于过滤文本节点。如果XML中没有其他内容,剩下的就是人员节点。
对于东西下的每个节点(姓名,电话,电子邮件,地区,城市)
el.getElementsByTagName("name").item(0).getTextContent();
el.getElementsByTagName("name")
将提取东西下的“名称”节点,
.item(0)
将为您提供第一个节点
并且.getTextContent()
将获取内部的文本内容。
修改强> 既然我们有杰克逊,我会以不同的方式做到这一点。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();
}
}
答案 1 :(得分:4)
public class XMLParser {
public static void main(String[] args){
try {
DocumentBuilder dBuilder = DocumentBuilderFactory.newInstance().newDocumentBuilder();
Document doc = dBuilder.parse(new File("xml input"));
NodeList nl=doc.getDocumentElement().getChildNodes();
for(int k=0;k<nl.getLength();k++){
printTags((Node)nl.item(k));
}
} catch (Exception e) {/*err handling*/}
}
public static void printTags(Node nodes){
if(nodes.hasChildNodes() || nodes.getNodeType()!=3){
System.out.println(nodes.getNodeName()+" : "+nodes.getTextContent());
NodeList nl=nodes.getChildNodes();
for(int j=0;j<nl.getLength();j++)printTags(nl.item(j));
}
}
}
递归循环并打印出文档中的所有xml子标记,以防您不必更改代码来处理xml中的动态更改,前提是它是一个格式良好的xml。
答案 2 :(得分:1)
这是使用JDOM循环遍历XML元素的另一种方法。
List<Element> nodeNodes = inputNode.getChildren();
if (nodeNodes != null) {
for (Element nodeNode : nodeNodes) {
List<Element> elements = nodeNode.getChildren(elementName);
if (elements != null) {
elements.size();
nodeNodes.removeAll(elements);
}
}