A simple way to read an XML file in Java

This is the simplest way to read data from an XML file into a Java program. I have also included some basic error checking, so you can directly cut-paste this code with a few changes ofcourse. All you have to do is change the XML tags within the program to match those that are present in your XML file.

XML File

 <book>
<person>
  <first>Kiran</first>
  <last>Pai</last>
  <age>22</age>
</person>
<person>
  <first>Bill</first>
  <last>Gates</last>
  <age>46</age>
</person>
<person>
  <first>Steve</first>
  <last>Jobs</last>
  <age>40</age>
</person>
</book>

Program Output

Root element of the doc is book
Total no of people : 3
First Name : Kiran
Last Name : Pai
Age : 22
First Name : Bill
Last Name : Gates
Age : 46
First Name : Steve
Last Name : Jobs
Age : 40

Program Listing

The Java program to read the above XML file is shown below. Go through the program twice and you will understand all its parts. It may look intimidating at first sight, but believe me its very simple.

import java.io.File;
import org.w3c.dom.Document;
import org.w3c.dom.*;

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException; 

public class ReadAndPrintXMLFile{

    public static void main (String argv []){
    try {

            DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
            Document doc = docBuilder.parse (new File("book.xml"));

            // normalize text representation
            doc.getDocumentElement ().normalize ();
            System.out.println ("Root element of the doc is " + 
                 doc.getDocumentElement().getNodeName());


            NodeList listOfPersons = doc.getElementsByTagName("person");
            int totalPersons = listOfPersons.getLength();
            System.out.println("Total no of people : " + totalPersons);

            for(int s=0; s<listOfPersons.getLength() ; s++){


                Node firstPersonNode = listOfPersons.item(s);
                if(firstPersonNode.getNodeType() == Node.ELEMENT_NODE){


                    Element firstPersonElement = (Element)firstPersonNode;

                    //-------
                    NodeList firstNameList = firstPersonElement.getElementsByTagName("first");
                    Element firstNameElement = (Element)firstNameList.item(0);

                    NodeList textFNList = firstNameElement.getChildNodes();
                    System.out.println("First Name : " + 
                           ((Node)textFNList.item(0)).getNodeValue().trim());

                    //-------
                    NodeList lastNameList = firstPersonElement.getElementsByTagName("last");
                    Element lastNameElement = (Element)lastNameList.item(0);

                    NodeList textLNList = lastNameElement.getChildNodes();
                    System.out.println("Last Name : " + 
                           ((Node)textLNList.item(0)).getNodeValue().trim());

                    //----
                    NodeList ageList = firstPersonElement.getElementsByTagName("age");
                    Element ageElement = (Element)ageList.item(0);

                    NodeList textAgeList = ageElement.getChildNodes();
                    System.out.println("Age : " + 
                           ((Node)textAgeList.item(0)).getNodeValue().trim());

                    //------


                }//end of if clause


            }//end of for loop with s var


        }catch (SAXParseException err) {
        System.out.println ("** Parsing error" + ", line " 
             + err.getLineNumber () + ", uri " + err.getSystemId ());
        System.out.println(" " + err.getMessage ());

        }catch (SAXException e) {
        Exception x = e.getException ();
        ((x == null) ? e : x).printStackTrace ();

        }catch (Throwable t) {
        t.printStackTrace ();
        }
        //System.exit (0);

    }//end of main


}

There are better implementations of reading XML files which would work for any XML file. The above one would require a few changes every time the XML tag names change. But this is much more simpler than the other programs.

You might also like...

Comments

Kiran Pai

Contribute

Why not write for us? Or you could submit an event or a user group in your area. Alternatively just tell us what you think!

Our tools

We've got automatic conversion tools to convert C# to VB.NET, VB.NET to C#. Also you can compress javascript and compress css and generate sql connection strings.

“Computer Science is no more about computers than astronomy is about telescopes.” - E. W. Dijkstra