Parsing / Reading XML file in Java.
- By Hanumant Shikhare on April 17, 2009
- How-To, Java
While working on a small requirement, I had to write a small piece of code that can parse an XML file in Java. Although there are number of libraries available in Java which does this task efficiently, I ended up in using normal Java XML parsing using org.w3c.dom parser.
You may want to copy this code and customize it to your to fit it for your need.
Following is the sample XML file that I used.
XML File
<?xml version="1.0"?> <students> <student> <name>John</name> <grade>B</grade> <age>12</age> </student> <student> <name>Mary</name> <grade>A</grade> <age>11</age> </student> <student> <name>Simon</name> <grade>A</grade> <age>18</age> </student> </students>
Source Code of XML Parser
Following is the source code of sample XML parser in Java that I used to parse sample XML file.
package net.viralpatel.java.xmlparser;
import java.io.File;
import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
public class XMLParser {
public void getAllUserNames(String fileName) {
try {
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
DocumentBuilder db = dbf.newDocumentBuilder();
File file = new File(fileName);
if (file.exists()) {
Document doc = db.parse(file);
Element docEle = doc.getDocumentElement();
// Print root element of the document
System.out.println("Root element of the document: "
+ docEle.getNodeName());
NodeList studentList = docEle.getElementsByTagName("student");
// Print total student elements in document
System.out
.println("Total students: " + studentList.getLength());
if (studentList != null && studentList.getLength() > 0) {
for (int i = 0; i < studentList.getLength(); i++) {
Node node = studentList.item(i);
if (node.getNodeType() == Node.ELEMENT_NODE) {
System.out
.println("=====================");
Element e = (Element) node;
NodeList nodeList = e.getElementsByTagName("name");
System.out.println("Name: "
+ nodeList.item(0).getChildNodes().item(0)
.getNodeValue());
nodeList = e.getElementsByTagName("grade");
System.out.println("Grade: "
+ nodeList.item(0).getChildNodes().item(0)
.getNodeValue());
nodeList = e.getElementsByTagName("age");
System.out.println("Age: "
+ nodeList.item(0).getChildNodes().item(0)
.getNodeValue());
}
}
} else {
System.exit(1);
}
}
} catch (Exception e) {
System.out.println(e);
}
}
public static void main(String[] args) {
XMLParser parser = new XMLParser();
parser.getAllUserNames("c:\\test.xml");
}
}
Following is the output of our XML parser.
Program Output
Root element of the document: students
Total students: 3
==============================
Name: John
Grade: B
Age: 12
==============================
Name: Mary
Grade: A
Age: 11
==============================
Name: Simon
Grade: A
Age: 18
You can customize this code for your input XML file.
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There is bug or bad display at line 33.
if (studentList != null &amp;amp;&amp;amp; studentList.getLength() > 0) {
Thanks JLemire, I have updated the source and removed the typo error.
Hi How to read this XML file and store it into HashSet?
You want want to look at vtd-xml, the latest and most advanced parsing/indexing/XPath engine that offers a lot of cool features
vtd-xml