Better way to parse xml

Question

I\'ve been parsing XML like this for years, and I have to admit when the number of different element becomes larger I find it a bit boring and exhausting to do, here is what I m

Answer 1

    import java.io.File;
import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.ArrayList;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathFactory;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;

public class JXML {
private DocumentBuilder builder;
private Document doc = null;
private DocumentBuilderFactory factory ;
private XPathExpression expr = null;
private XPathFactory xFactory;
private XPath xpath;
private String xmlFile;
public static ArrayList XMLVALUE ;  


public JXML(String xmlFile){
    this.xmlFile = xmlFile;
}


private void xmlFileSettings(){     
    try {
        factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true);
        xFactory = XPathFactory.newInstance();
        xpath = xFactory.newXPath();
        builder = factory.newDocumentBuilder();
        doc = builder.parse(xmlFile);
    }
    catch (Exception e){
        System.out.println(e);
    }       
}



public String[] selectQuery(String query){
    xmlFileSettings();
    ArrayList records = new ArrayList();
    try {
        expr = xpath.compile(query);
        Object result = expr.evaluate(doc, XPathConstants.NODESET);
        NodeList nodes = (NodeList) result;
        for (int i=0; i

}