How to modify XML FILE IN JAVA

In this example, we demonstrate the use of DOM parser to modify an existing XML file :

1. Add a new element
2. Update existing element attribute
3. Update existing element value
4. Delete existing element

1. XML file

See before and after XML file.
File : file.xml – Original XML file.

<?xml version="1.0" encoding="UTF-8" standalone="no" ?> 
<company>
   <staff id="1">
	<firstname>yong</firstname>
	<lastname>mook kim</lastname>
	<nickname>mkyong</nickname>
	<salary>100000</salary>
   </staff>
</company>

Later, update above XML file via DOM XML Parser.

1. Update the staff attribute id = 2
2. Update salary value to 200000
3. Append a new “age” element under staff
4. Delete “firstname” element under staff

File : file.xml – Newly modified XML file.

<?xml version="1.0" encoding="UTF-8" standalone="no" ?> 
<company>
   <staff id="2">
	<lastname>mook kim</lastname>
	<nickname>mkyong</nickname>
	<salary>2000000</salary> 
        <age>28</age> 
   </staff>
</company>

2. DOM Parser

DOM XML parser to update an above XML file.

import java.io.File;
import java.io.IOException;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.transform.Transformer;
import javax.xml.transform.TransformerException;
import javax.xml.transform.TransformerFactory;
import javax.xml.transform.dom.DOMSource;
import javax.xml.transform.stream.StreamResult;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
 
public class ModifyXMLFile {
 
	public static void main(String argv[]) {
 
	   try {
		String filepath = "c:\\file.xml";
		DocumentBuilderFactory docFactory = DocumentBuilderFactory.newInstance();
		DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
		Document doc = docBuilder.parse(filepath);
 
		// Get the root element
		Node company = doc.getFirstChild();
 
		// Get the staff element , it may not working if tag has spaces, or
		// whatever weird characters in front...it's better to use
		// getElementsByTagName() to get it directly.
		// Node staff = company.getFirstChild();
 
		// Get the staff element by tag name directly
		Node staff = doc.getElementsByTagName("staff").item(0);
 
		// update staff attribute
		NamedNodeMap attr = staff.getAttributes();
		Node nodeAttr = attr.getNamedItem("id");
		nodeAttr.setTextContent("2");
 
		// append a new node to staff
		Element age = doc.createElement("age");
		age.appendChild(doc.createTextNode("28"));
		staff.appendChild(age);
 
		// loop the staff child node
		NodeList list = staff.getChildNodes();
 
		for (int i = 0; i < list.getLength(); i++) {
 
                   Node node = list.item(i);
 
		   // get the salary element, and update the value
		   if ("salary".equals(node.getNodeName())) {
			node.setTextContent("2000000");
		   }
 
                   //remove firstname
		   if ("firstname".equals(node.getNodeName())) {
			staff.removeChild(node);
		   }
 
		}
 
		// write the content into xml file
		TransformerFactory transformerFactory = TransformerFactory.newInstance();
		Transformer transformer = transformerFactory.newTransformer();
		DOMSource source = new DOMSource(doc);
		StreamResult result = new StreamResult(new File(filepath));
		transformer.transform(source, result);
 
		System.out.println("Done");
 
	   } catch (ParserConfigurationException pce) {
		pce.printStackTrace();
	   } catch (TransformerException tfe) {
		tfe.printStackTrace();
	   } catch (IOException ioe) {
		ioe.printStackTrace();
	   } catch (SAXException sae) {
		sae.printStackTrace();
	   }
	}
}

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Read XML file in java

In this tutorial, we will show you how to read an XML file via DOM XML parser. DOM parser parses the entire XML document and loads it into memory; then models it in a “TREE” structure for easy traversal or manipulation.
In short, it turns a XML file into DOM or Tree structure, and you have to traverse a node by node to get what you want.

What is Node?
In the DOM, everything in an XML document is a node, read this.

Warning
DOM Parser is slow and consumes a lot of memory when it loads an XML document which contains a lot of data. Please consider SAX parser as solution for it, SAX is faster than DOM and use less memory.

1. DOM XML Parser Example

This example shows you how to get the node by “name”, and display the value.

/Users/mkyong/staff.xml

<?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>

ReadXMLFile.java

package com.mkyong.seo;
 
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();
    }
  }
 
}

Result

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

2. Looping the Node

This example reads the same “staff.xml“, and showing you how to loop the node one by one, and print out the node name and value, and also the attribute if any.

ReadXMLFile2.java

package com.mkyong.seo;
 
import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.NamedNodeMap;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
 
public class ReadXMLFile2 {
 
  public static void main(String[] args) {
 
    try {
 
	File file = new File("/Users/mkyong/staff.xml");
 
	DocumentBuilder dBuilder = DocumentBuilderFactory.newInstance()
                             .newDocumentBuilder();
 
	Document doc = dBuilder.parse(file);
 
	System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
 
	if (doc.hasChildNodes()) {
 
		printNote(doc.getChildNodes());
 
	}
 
    } catch (Exception e) {
	System.out.println(e.getMessage());
    }
 
  }
 
  private static void printNote(NodeList nodeList) {
 
    for (int count = 0; count < nodeList.getLength(); count++) {
 
	Node tempNode = nodeList.item(count);
 
	// make sure it's element node.
	if (tempNode.getNodeType() == Node.ELEMENT_NODE) {
 
		// get node name and value
		System.out.println("\nNode Name =" + tempNode.getNodeName() + " [OPEN]");
		System.out.println("Node Value =" + tempNode.getTextContent());
 
		if (tempNode.hasAttributes()) {
 
			// get attributes names and values
			NamedNodeMap nodeMap = tempNode.getAttributes();
 
			for (int i = 0; i < nodeMap.getLength(); i++) {
 
				Node node = nodeMap.item(i);
				System.out.println("attr name : " + node.getNodeName());
				System.out.println("attr value : " + node.getNodeValue());
 
			}
 
		}
 
		if (tempNode.hasChildNodes()) {
 
			// loop again if has child nodes
			printNote(tempNode.getChildNodes());
 
		}
 
		System.out.println("Node Name =" + tempNode.getNodeName() + " [CLOSE]");
 
	}
 
    }
 
  }
 
}

Result :

Root element :company
 
Node Name =company [OPEN]
Node Value =
 
		yong
		mook kim
		mkyong
		100000
 
 
		low
		yin fong
		fong fong
		200000
 
 
 
Node Name =staff [OPEN]
Node Value =
		yong
		mook kim
		mkyong
		100000
 
attr name : id
attr value : 1001
 
Node Name =firstname [OPEN]
Node Value =yong
Node Name =firstname [CLOSE]
 
Node Name =lastname [OPEN]
Node Value =mook kim
Node Name =lastname [CLOSE]
 
Node Name =nickname [OPEN]
Node Value =mkyong
Node Name =nickname [CLOSE]
 
Node Name =salary [OPEN]
Node Value =100000
Node Name =salary [CLOSE]
Node Name =staff [CLOSE]
 
Node Name =staff [OPEN]
Node Value =
		low
		yin fong
		fong fong
		200000
 
attr name : id
attr value : 2001
 
Node Name =firstname [OPEN]
Node Value =low
Node Name =firstname [CLOSE]
 
Node Name =lastname [OPEN]
Node Value =yin fong
Node Name =lastname [CLOSE]
 
Node Name =nickname [OPEN]
Node Value =fong fong
Node Name =nickname [CLOSE]
 
Node Name =salary [OPEN]
Node Value =200000
Node Name =salary [CLOSE]
Node Name =staff [CLOSE]
Node Name =company [CLOSE]

Note
You may interest at this How to get Alexa Ranking In Java. It shows you how to use DOM to parse the Alexa XML result.

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