XML parsing returns null

Question

I am parsing the below XML using the Dom Parser.

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<company>
    <Staff id="1">
        <firstname>Achyut</firstname>
        <lastname>khanna</lastname>
        <nickname>Achyut</nickname>
        <salary>900000</salary>
    </Staff>
</company>

If I only need firstName from the XML why is returning null ?

private String getNodeValue(Node node) {
        Node nd = node.getFirstChild();     
        try {
            if (nd == null) {
                return node.getNodeValue();             
            }
            else {              
                 getNodeValue(nd);
            }
        } catch (Exception e) {

            e.printStackTrace();
        }
        return null;
    }

What XML library are you using? What kind of object (what package & class) is the Node argument? — Stewart
– Stewart, Commented Sep 2, 2013 at 9:44
Apart from the xml-libs : you must return the value of your recursive call, too. return getNodeValue(nd) for example — Jan Piel
– Jan Piel, Commented Sep 2, 2013 at 9:53
@Jan Piel thanks for the comment ,I forget a return statement. Thanks for reply !! — Achyut
– Achyut, Commented Sep 2, 2013 at 10:02

AurA · Accepted Answer · 2013-09-02 09:47:04Z

1

You must fetch the nodelist and then pass the appropriate Node value as parameter when you call your defined function.

NodeList n = item.getElementsByTagName("Staff");

Then call your function

String firstName = getNodeValue(n.item(0));

answered Sep 2, 2013 at 9:47

AurA

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2 Comments

Achyut Over a year ago

but using debugger I got that the it comes to return line and show its value to Achyut but again enters inside else.Why ?

AurA Over a year ago

the code you wrote is a function which returns the first child of the passed Node value if it exists, here the only issue might be you are calling it with the wrong node value.

dharam · Accepted Answer · 2013-09-02 09:46:03Z

First of all, I do not recommend parsing XMLs using DOM traversals. I would recommend you to use the OXMs (JaxB or XMLbeans). But still if you are interested in doing it this way:

Here is the code

public class T2 {
public static void main(String []args) throws ParserConfigurationException, SAXException, IOException{
DocumentBuilder db = null;

String xmlString = "<?xml version='1.0' encoding='UTF-8' standalone='no'?><company>    <Staff id='1'>        <firstname>Achyut</firstname>        <lastname>khanna</lastname>        <nickname>Achyut</nickname>        <salary>900000</salary>    </Staff></company>";
Document doc = null;
InputSource is = new InputSource();

is.setCharacterStream(new StringReader(xmlString));

    db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
    doc = db.parse(is);

    NodeList nodes = doc.getElementsByTagName("firstname");

    for (int i = 0; i < nodes.getLength(); i++) {
        if (nodes.item(i) instanceof Element) {
             Node node = (Node) nodes.item(i);
            nodes.item(i);

            String fName  = getCharacterDataFromElement(node);
            System.out.println(fName);
        }
    }


}
private static String getCharacterDataFromElement(Node e) {
    Node child = e.getFirstChild();
    if (child instanceof CharacterData) {
        CharacterData cd = (CharacterData) child;
        return cd.getData();
    }
    return null;
}
}

The code above prints Achyut

forty-two · Accepted Answer · 2013-09-02 11:59:52Z

0

Save yourself a lot of code by using XPath instead:

XPath xp = XPathFactory.newInstance().newXPath();
InputSource in = new InputSource(...);
String fn = xp.evaluate("/company/Staff[@id='1']/firstname/text()", in);
System.out.println(fn);

printts:

Achyut

answered Sep 2, 2013 at 11:59

forty-two

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