java 使用dom3c解析xml文件问题，多级节点解析

Question

读取xml文件，
获取如下标签的NodeList，

NodeList impList = doc.getElementsByTagName("IMPLEMENTATION-DATA-TYPE");
for(int i = 0; i < impList.getLength(); i++) {
    Element impElement = (Element) impList.item(i); 
    NodeList impSubList = impElement.getElementsByTagName("IMPLEMENTATION-DATA-TYPE-ELEMENT");
    System.out.println("impSubList legth= " + impSubList.getLength());
    }

我想做到输出的impSubList legth=5，但是实际输出的长度是这个节点下所有的这个便签的长度。怎么一级一级解析下去 获取长度？

Answer 1

直接上代码

package cn.next;
import java.io.IOException;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import org.w3c.dom.Document;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
import org.w3c.dom.Text;

public class Test {
    public static void ReadTreeStructure(NodeList nodes) {

        for (int i = 0; i < nodes.getLength(); i++) {
            if (nodes.item(i).getNodeType() == Node.ELEMENT_NODE) {
                System.out.print(nodes.item(i).getNodeName() + " ");
                String value = ((Text) (nodes.item(i).getFirstChild())).getData().trim();
                if (value.getBytes().length != 0) {
                    System.out.print(value);
                }
                System.out.println();
            }

                        //如果节点还有子节点,递归调用
            if (nodes.item(i).getChildNodes().getLength() != 0) {
                ReadTreeStructure(nodes.item(i).getChildNodes());
            }
        }
    }

    public static void main(String[] args) {

        DocumentBuilderFactory docbf = DocumentBuilderFactory.newInstance();
        try {
            DocumentBuilder docb = docbf.newDocumentBuilder();
            Document doc = docb.parse("***.xml");
            NodeList nodes = doc.getElementsByTagName("IMPLEMENTATION-DATA-TYPE");
            ReadTreeStructure(nodes);
        } catch (ParserConfigurationException e) {
            e.printStackTrace();
        } catch (SAXException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }

}

Answer 2

设置一个count变量 进行判断 然后累加count

Answer 3

package com.test;

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

public class DomXML {

    public static void main(String[] args) {
        try {
            File file = new File("e:/People.xml");  
            DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();   
            DocumentBuilder builder = factory.newDocumentBuilder();   
            Document document = builder.parse(file);   
            Element element = document.getDocumentElement();

            List<People> peopleList = new ArrayList<People>();
            NodeList peopleNodes = element.getElementsByTagName("People"); 
            for(int i=0;i<peopleNodes.getLength();i++){   
                People people = new People();
                Element peopleElement = (Element) peopleNodes.item(i);
                people.setId(peopleElement.getAttribute("id"));
                NodeList childPeopleNodes = peopleElement.getChildNodes();
                for(int j=0;j<childPeopleNodes.getLength();j++){
                    //DOM解析时候注意子节点前面的空格也会被解析
                    if(childPeopleNodes.item(j) instanceof Element){
                        Element childPeopleElement = (Element) childPeopleNodes.item(j);
                        if(childPeopleElement.getNodeType()==Node.ELEMENT_NODE){  
                            if(childPeopleElement.getNodeName().equals("Name")){
                                people.setEnglishName(childPeopleElement.getAttribute("en"));
                                people.setName(childPeopleElement.getTextContent());
                            }
                            else if(childPeopleElement.getNodeName().equals("Age")){
                                people.setAge(childPeopleElement.getTextContent());    
                            }
                        }
                    }
                }
                peopleList.add(people);
            }

            for(People people : peopleList){
                System.out.println(people.getId()+"\t"+people.getName()+"\t"+people.getEnglishName()+"\t"+people.getAge());
            }

        } catch (Exception e) {
            // TODO 自动生成的 catch 块
            e.printStackTrace();
        } 


    }
}

Answer 4

要不用dom4j解析吧，可以直接拿子节点的列表，而不是所有的子孙节点