Java HTML / XML How to - Parse simple XML file with Java and SAX

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Question

We would like to know how to parse simple XML file with Java and SAX.

Answer

import java.io.File;
//  w w  w .  j a  v a  2s  .c o m
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 Main {

  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);
        }
      }
    });
  }
}

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