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Java - XPath  Tutorial

                         

Showing all elements of an XML file using Java xpath

In this example we have created an XML file "person.xml" first, which is necessary to execute XPath query on to it. This "persons.xml" contains information related to name, age, gender of different persons.

Here is the full source code for persons.xml file as follows :

persons.xml

<?xml version="1.0" ?>
<information>
    <person id="1">
        <name>Deep</name>
        <age>34</age>
        <gender>Male</gender>
    </person>
 
     <person id="2">
        <name>Kumar</name>
        <age>24</age>
        <gender>Male</gender>
    </person>
 
    <person id="3">
        <name>Deepali</name>
        <age>19</age>
        <gender>Female</gender>
    </person>

    <!-- more persons... -->
</information>

Now we have declared a class XPathDemo  and in this class we are parsing the XML file with JAXP. First of all we do need to load the document into DOM Document object. We have placed that persons.xml file in that current working directory.

    DocumentBuilderFactory domFactory = 
    DocumentBuilderFactory.newInstance();

          domFactory.setNamespaceAware(true); 
    DocumentBuilder builder = domFactory.newDocumentBuilder();
    Document doc = builder.parse("persons.xml");

Above lines of code parses "persons.xml" file and creates a Document object. Next we have created XPath object with the use of XPathFactory.

XPath xpath = XPathFactory.newInstance().newXPath();

Now we compile the path with the use of compile() method. Finally we will evaluate XPath expression and then we can print all those elements.
Here is the example code for XPathDemo.java as follows:

XPathDemo.java

import org.w3c.dom.*;
import javax.xml.xpath.*;
import javax.xml.parsers.*;
import java.io.IOException;
import org.xml.sax.SAXException;

public class XPathDemo {

  public static void main(String[] args
   throws ParserConfigurationException, SAXException, 
          IOException, XPathExpressionException {

    DocumentBuilderFactory domFactory = 
    DocumentBuilderFactory.newInstance
();
          domFactory.setNamespaceAware(true)
    DocumentBuilder builder = domFactory.newDocumentBuilder();
    Document doc = builder.parse("persons.xml");
    XPath xpath = XPathFactory.newInstance().newXPath();
       // XPath Query for showing all nodes value
    XPathExpression expr = xpath.compile("//person/*/text()");

    Object result = expr.evaluate(doc, XPathConstants.NODESET);
    NodeList nodes = (NodeListresult;
    for (int i = 0; i < nodes.getLength(); i++) {
     System.out.println(nodes.item(i).getNodeValue())
    }
  }
}

To run this example follow these steps as mention below :

  1. First create and save an XML file "persons.xml"
  2. Create XPathDemo.java and compile it.
  3. Execute XPathDemo you will get the following output on your console

Output:

Download Source Code

                         

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