Create XML document using nodeList

nath picture nath · Apr 26, 2011 · Viewed 44.6k times · Source

I need to create a XML Document object using the NodeList. Can someone pls help me to do this. This is my Java code:

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.*; 
import org.w3c.dom.*;

public class ReadFile {

    public static void main(String[] args) {
        String exp = "/configs/markets";
        String path = "testConfig.xml";
        try {
            Document xmlDocument = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(path);
            XPath xPath = XPathFactory.newInstance().newXPath();
            XPathExpression xPathExpression = xPath.compile(exp);
            NodeList nodes = (NodeList)
              xPathExpression.evaluate(xmlDocument,
                                       XPathConstants.NODESET);

        } catch (Exception ex) {
            ex.printStackTrace();
        } 
    }
}

I want to have an XML file like this:

<configs>
    <markets>   
        <market>
            <name>Real</name>
        </market>
        <market>
            <name>play</name>
        </market>
    </markets>
</configs>

Thanks in advance.

Answer

MarcoS picture MarcoS · Apr 26, 2011

You should do it like this:

  • you create a new org.w3c.dom.Document newXmlDoc where you store the nodes in your NodeList,
  • you create a new root element, and append it to newXmlDoc
  • then, for each node n in your NodeList, you import n in newXmlDoc, and then you append n as a child of root

Here is the code:

public static void main(String[] args) {
    String exp = "/configs/markets/market";
    String path = "src/a/testConfig.xml";
    try {
        Document xmlDocument = DocumentBuilderFactory.newInstance()
                .newDocumentBuilder().parse(path);

        XPath xPath = XPathFactory.newInstance().newXPath();
        XPathExpression xPathExpression = xPath.compile(exp);
        NodeList nodes = (NodeList) xPathExpression.
                evaluate(xmlDocument, XPathConstants.NODESET);

        Document newXmlDocument = DocumentBuilderFactory.newInstance()
                .newDocumentBuilder().newDocument();
        Element root = newXmlDocument.createElement("root");
        newXmlDocument.appendChild(root);
        for (int i = 0; i < nodes.getLength(); i++) {
            Node node = nodes.item(i);
            Node copyNode = newXmlDocument.importNode(node, true);
            root.appendChild(copyNode);
        }

        printTree(newXmlDocument);
    } catch (Exception ex) {
        ex.printStackTrace();
    }
}

public static void printXmlDocument(Document document) {
    DOMImplementationLS domImplementationLS = 
        (DOMImplementationLS) document.getImplementation();
    LSSerializer lsSerializer = 
        domImplementationLS.createLSSerializer();
    String string = lsSerializer.writeToString(document);
    System.out.println(string);
}

The output is:

<?xml version="1.0" encoding="UTF-16"?>
<root><market>
            <name>Real</name>
        </market><market>
            <name>play</name>
        </market></root>

Some notes:

  • I've changed exp to /configs/markets/market, because I suspect you want to copy the market elements, rather than the single markets element
  • for the printXmlDocument, I've used the interesting code in this answer

I hope this helps.


If you don't want to create a new root element, then you may use your original XPath expression, which returns a NodeList consisting of a single node (keep in mind that your XML must have a single root element) that you can directly add to your new XML document.

See following code, where I commented lines from the code above:

public static void main(String[] args) {
    //String exp = "/configs/markets/market/";
    String exp = "/configs/markets";
    String path = "src/a/testConfig.xml";
    try {
        Document xmlDocument = DocumentBuilderFactory.newInstance()
                .newDocumentBuilder().parse(path);

        XPath xPath = XPathFactory.newInstance().newXPath();
        XPathExpression xPathExpression = xPath.compile(exp);
        NodeList nodes = (NodeList) xPathExpression.
        evaluate(xmlDocument,XPathConstants.NODESET);

        Document newXmlDocument = DocumentBuilderFactory.newInstance()
                .newDocumentBuilder().newDocument();
        //Element root = newXmlDocument.createElement("root");
        //newXmlDocument.appendChild(root);
        for (int i = 0; i < nodes.getLength(); i++) {
            Node node = nodes.item(i);
            Node copyNode = newXmlDocument.importNode(node, true);
            newXmlDocument.appendChild(copyNode);
            //root.appendChild(copyNode);
        }

        printXmlDocument(newXmlDocument);
    } catch (Exception ex) {
        ex.printStackTrace();
    }
}

This will give you the following output:

<?xml version="1.0" encoding="UTF-16"?>
<markets>   
        <market>
            <name>Real</name>
        </market>
        <market>
            <name>play</name>
        </market>
    </markets>