如何使用Java和SAX解析我的简单XML文件？

这是一个工作示例，尽管我建议您改用StAX，但您会发现SAX不太方便

import java.io.File;import javax.xml.parsers.SAXParser;import javax.xml.parsers.SAXParserFactory;import org.xml.sax.Attributes;import org.xml.sax.SAXException;import org.xml.sax.helpers.DefaultHandler;public class SAX2 {    public static void main(String[] args) throws Exception {        SAXParser parser = SAXParserFactory.newInstance().newSAXParser();        parser.parse(new File("test.xml"), new DefaultHandler() { @Override public void startElement(String uri, String localName,         String qName, Attributes atts) throws SAXException {     if (qName.equals("passenger")) {         System.out.println("id = " + atts.getValue(0));     } } @Override public void endElement(String uri, String localName, String qName)         throws SAXException { } @Override public void characters(char[] ch, int start, int length)         throws SAXException {     String text = new String(ch, start, length);     if (!text.trim().isEmpty()) {         System.out.println("name " + text);     } }        });    }}

输出

id = 001name Tom Cruiseid = 002name Tom Hanks