Read xml in java - all nodes and data sort

Question

Now my software will show me (Root element: data) from xml file, but I need list of nodes. This what Xml file have got for example:

(data have got "count" and "person")

(count = 1)

(person have got "name" and "cars")

(name = "Adam")

(cars have got "Porshe" and "minimini")

Softweare

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 void Ada()
{
File fXmlFile = new File("c:\\file.xml");
            DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
            Document doc = dBuilder.parse(fXmlFile);
            doc.getDocumentElement().normalize();

            System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
}

XML File:

<data>
    <count>1</count>
    <person>
        <name>Adam<name>
        <cars>
            <fast>Porshe</fast>
            <slow>MiniMini</slow>
        </cars>
    </person>
</data>

Answer 1

You can use:

doc.getChildNodes()

to get deeper and the recursively drill dow until there are no children left.

It will return a NodeList you can the use the size() to 'power' a for loop.

Answer 2

Take a look at XPath: http://developer.android.com/reference/javax/xml/xpath/package-summary.html

You can basically do anything you need with XML on Android using that.

So something like:

XPath x = XPathFactory.newInstance().newXPath();
String expression = "/data";
InputSource source = new InputSource("someXML.xml");
NodeSet nodes = (NodeSet)xpath.evaluate(expression, source, XPathConstants.NODESET);

That should give you a NodeSet of the nodes.

Hope that helps.