Java HTML / XML How to - Handle error for DOM parser








Question

We would like to know how to handle error for DOM parser.

Answer

import java.io.IOException;
import java.io.StringReader;
/*from w w  w . jav  a2s.  c  om*/
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;

import org.w3c.dom.Document;
import org.xml.sax.ErrorHandler;
import org.xml.sax.InputSource;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException;

public class Main {
  public static String getXMLData() {
    return "<x></a>";
  }

  static public void main(String[] arg) {
    boolean validate = true;

    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    dbf.setValidating(validate);
    dbf.setNamespaceAware(true);

    try {
      DocumentBuilder builder = dbf.newDocumentBuilder();
      builder.setErrorHandler(new MyErrorHandler());
      InputSource is = new InputSource(new StringReader(getXMLData()));
      Document doc = builder.parse(is);
    } catch (SAXException e) {
      System.out.println(e);
    } catch (ParserConfigurationException e) {
      System.err.println(e);
    } catch (IOException e) {
      System.err.println(e);
    }
  }
}

class MyErrorHandler implements ErrorHandler {
  public void warning(SAXParseException e) throws SAXException {
    show("Warning", e);
    throw (e);
  }

  public void error(SAXParseException e) throws SAXException {
    show("Error", e);
    throw (e);
  }

  public void fatalError(SAXParseException e) throws SAXException {
    show("Fatal Error", e);
    throw (e);
  }

  private void show(String type, SAXParseException e) {
    System.out.println(type + ": " + e.getMessage());
    System.out.println("Line " + e.getLineNumber() + " Column "
        + e.getColumnNumber());
    System.out.println("System ID: " + e.getSystemId());
  }
}