Java XPath Example Tutorial

Welcome to Java XPath Example Tutorial. XPath provides syntax to define part of an XML document. XPath Expression is a query language to select part of the XML document based on the query String. Using XPath Expressions, we can find nodes in any xml document satisfying the query string.

Java XPath

javax.xml.xpath package provides XPath support in Java. To create XPathExpression, XPath API provide factory methods.

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XPathFactory xpathFactory = XPathFactory.newInstance();
XPath xpath = xpathFactory.newXPath();
XPathExpression expr = xpath.compile(XPATH_EXPRESSION_STRING);
Object result = expr.evaluate(Object item, QName returnType);

Java XPath supported return types are defined in XPathConstants class.

1. XPathConstants.STRING
2. XPathConstants.NUMBER
3. XPathConstants.BOOLEAN
4. XPathConstants.NODE
5. XPathConstants.NODESET

Java XPath Example

In this java XPath example tutorial, we have following sample xml file.

employees.xml

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<?xml version="1.0" encoding="UTF-8"?>
<Employees>
	<Employee id="1">
		<age>29</age>
		<name>Pankaj</name>
		<gender>Male</gender>
		<role>Java Developer</role>
	</Employee>
	<Employee id="2">
		<age>35</age>
		<name>Lisa</name>
		<gender>Female</gender>
		<role>CEO</role>
	</Employee>
	<Employee id="3">
		<age>40</age>
		<name>Tom</name>
		<gender>Male</gender>
		<role>Manager</role>
	</Employee>
	<Employee id="4">
		<age>25</age>
		<name>Meghna</name>
		<gender>Female</gender>
		<role>Manager</role>
	</Employee>
</Employees>

We will implement following functions in our Java XPath example program.

1. A function that will return the Employee Name for input ID.
2. Return list of Employees Name with age greater than the input age.
3. Return list of Female Employees Name.

Here is the final implementation class for Java XPath example program.

XPathQueryExample.java

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package com.journaldev.xml;


import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import javax.xml.xpath.XPath;
import javax.xml.xpath.XPathConstants;
import javax.xml.xpath.XPathExpression;
import javax.xml.xpath.XPathExpressionException;
import javax.xml.xpath.XPathFactory;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;


public class XPathQueryExample {

    public static void main(String[] args) {
        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        factory.setNamespaceAware(true);
        DocumentBuilder builder;
        Document doc = null;
        try {
            builder = factory.newDocumentBuilder();
            doc = builder.parse("/Users/pankaj/employees.xml");

            // Create XPathFactory object
            XPathFactory xpathFactory = XPathFactory.newInstance();

            // Create XPath object
            XPath xpath = xpathFactory.newXPath();

            String name = getEmployeeNameById(doc, xpath, 4);
            System.out.println("Employee Name with ID 4: " + name);

            List<String> names = getEmployeeNameWithAge(doc, xpath, 30);
            System.out.println("Employees with 'age>30' are:" + Arrays.toString(names.toArray()));

            List<String> femaleEmps = getFemaleEmployeesName(doc, xpath);
            System.out.println("Female Employees names are:" +
                    Arrays.toString(femaleEmps.toArray()));

        } catch (ParserConfigurationException | SAXException | IOException e) {
            e.printStackTrace();
        }

    }


    private static List<String> getFemaleEmployeesName(Document doc, XPath xpath) {
        List<String> list = new ArrayList<>();
        try {
            //create XPathExpression object
            XPathExpression expr =
                xpath.compile("/Employees/Employee[gender='Female']/name/text()");
            //evaluate expression result on XML document
            NodeList nodes = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
            for (int i = 0; i < nodes.getLength(); i++)
                list.add(nodes.item(i).getNodeValue());
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }
        return list;
    }


    private static List<String> getEmployeeNameWithAge(Document doc, XPath xpath, int age) {
        List<String> list = new ArrayList<>();
        try {
            XPathExpression expr =
                xpath.compile("/Employees/Employee[age>" + age + "]/name/text()");
            NodeList nodes = (NodeList) expr.evaluate(doc, XPathConstants.NODESET);
            for (int i = 0; i < nodes.getLength(); i++)
                list.add(nodes.item(i).getNodeValue());
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }
        return list;
    }


    private static String getEmployeeNameById(Document doc, XPath xpath, int id) {
        String name = null;
        try {
            XPathExpression expr =
                xpath.compile("/Employees/Employee[@id='" + id + "']/name/text()");
            name = (String) expr.evaluate(doc, XPathConstants.STRING);
        } catch (XPathExpressionException e) {
            e.printStackTrace();
        }

        return name;
    }

}

When we run above java XPath example program, it results in following output.

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Employee Name with ID 4: Meghna
Employees with 'age>30' are:[Lisa, Tom]
Female Employees names are:[Lisa, Meghna]

Notice that first few lines are to read the XML file as Document. Then we are reusing the Document and XPath object in all the methods. Above program shows example of NODESET and STRING as result Object.

That’s all for a quick roundup on Java XPath.

About Pankaj

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