What is the best/simplest way to read in an XML file in Java application? [closed]

Tags: java xml file
By : rmcc
Source: Stackoverflow.com
Question!

Currently our Java application uses the values held within a tab delimited *.cfg file. We need to change this application so that it now uses an XML file.

What is the best/simplest library to use in order to read in values from this file?

By : rmcc


Answers

This is what I use. http://marketmovers.blogspot.com/2014/02/the-easy-way-to-read-xml-in-java.html It sits on top of the standard JDK tools, so if it's missing some feature you can always use the JDK version.

This really makes things easier for me. It's especially nice when I'm reading a config file that was saved by and older version of the software, or was manually edited by a user. It's very robust and won't throw an exception if some data is not exactly in the format you expect.



<?xml version="1.0"?>
<company>
    <staff id="1001">
        <firstname>yong</firstname>
        <lastname>mook kim</lastname>
        <nickname>mkyong</nickname>
        <salary>100000</salary>
    </staff>
    <staff id="2001">
        <firstname>low</firstname>
        <lastname>yin fong</lastname>
        <nickname>fong fong</nickname>
        <salary>200000</salary>
    </staff>
</company>



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

public class ReadXMLFile {

  public static void main(String argv[]) {

    try {

    File fXmlFile = new File("/Users/mkyong/staff.xml");
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
    Document doc = dBuilder.parse(fXmlFile);

    //optional, but recommended
    //read this - http://stackoverflow.com/questions/13786607/normalization-in-dom-parsing-with-java-how-does-it-work
    doc.getDocumentElement().normalize();

    System.out.println("Root element :" + doc.getDocumentElement().getNodeName());

    NodeList nList = doc.getElementsByTagName("staff");

    System.out.println("----------------------------");

    for (int temp = 0; temp < nList.getLength(); temp++) {

        Node nNode = nList.item(temp);

        System.out.println("\nCurrent Element :" + nNode.getNodeName());

        if (nNode.getNodeType() == Node.ELEMENT_NODE) {

            Element eElement = (Element) nNode;

            System.out.println("Staff id : " + eElement.getAttribute("id"));
            System.out.println("First Name : " + eElement.getElementsByTagName("firstname").item(0).getTextContent());
            System.out.println("Last Name : " + eElement.getElementsByTagName("lastname").item(0).getTextContent());
            System.out.println("Nick Name : " + eElement.getElementsByTagName("nickname").item(0).getTextContent());
            System.out.println("Salary : " + eElement.getElementsByTagName("salary").item(0).getTextContent());

        }
    }
    } catch (Exception e) {
    e.printStackTrace();
    }
  }

}


----------------

Root element :company
----------------------------

Current Element :staff
Staff id : 1001
First Name : yong
Last Name : mook kim
Nick Name : mkyong
Salary : 100000

Current Element :staff
Staff id : 2001
First Name : low
Last Name : yin fong
Nick Name : fong fong
Salary : 200000
By : ran


Here's a really simple API that I created for reading simple XML files in Java. It's incredibly simple and easy to use. Hope it's useful for you.

http://argonrain.wordpress.com/2009/10/27/000/

By : Chris


This video can help you solving your question :)
By: admin