Java string to xml to list

Question

I'm new to Java and need some help. I have a XML that looks like:

String pXML =
"<root>
     <x>1</x>
     <x>2</x>
     <x>3</x>
     <x>4</x>
 </root>"

And I would like to get a List object that contains all of the values inside x tag.

I've tried with javax.xml.parsers.DocumentBuilderFactory:

DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance(); 
DocumentBuilder builder = factory.newDocumentBuilder();
document = (Document) builder.parse( new InputSource( new StringReader(pXML) ) );
Node n = document.getFirstChild();
NodeList n1 = n.getChildNodes();
//and then I go through all the nodes and insert the values into a list

But this doesn't contain the x nodes.

I suspect the quote marks are causing the parser to skip all of the XML. — Gilbert Le Blanc
– Gilbert Le Blanc, Commented Mar 26, 2013 at 13:47
No, the pXML object I get as an input parametar, this is for you to understand what the xml looks like... — informerica
– informerica, Commented Mar 26, 2013 at 13:57

Boris the Spider · Accepted Answer · 2013-03-26 14:52:12Z

3

You can use XPath to get the values of all the x nodes as follows:

public static void main(String[] args) throws SAXException, ParserConfigurationException, IOException, XPathExpressionException {
    final String pXML = "<root><x>1</x><x>2</x><x>3</x><x>4</x></root>";
    final Document document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(new ByteArrayInputStream(pXML.getBytes()));
    final XPathExpression xPathExpression = XPathFactory.newInstance().newXPath().compile("//x/text()");
    final NodeList nodeList = (NodeList) xPathExpression.evaluate(document, XPathConstants.NODESET);
    final List<String> values = new LinkedList<>();
    for (int i = 0; i < nodeList.getLength(); ++i) {
        values.add(nodeList.item(i).getNodeValue());
    }
    System.out.println(values);
}

Output:

[1, 2, 3, 4]

This has the advantage of being a very general solution, easily adaptable if the structure of the XML changes.

It also has the advantage of being, in my opinion, much more understandable than iterating over the nodes in the Document by hand.

edited Mar 26, 2013 at 14:52

answered Mar 26, 2013 at 13:51

Boris the Spider

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2 Comments

Bruno Grieder Over a year ago

Agreed and using xPath is much less error prone. Iterating child nodes of an element may give you a blend a Text nodes and (sub) Elements that you would not necessarily expect.

informerica Over a year ago

Figured out what was the problem - in my code NodeValues in NodeList had non-trimmed values, so it had tons of values, from '\n' and ' '. Just a mess of chars and strings...

Bruno Grieder · Accepted Answer · 2013-03-26 13:46:34Z

0

Try Node n = document.getDocumentElement(); to recover the root element of your XML

answered Mar 26, 2013 at 13:46

Bruno Grieder

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Comments

SK. · Accepted Answer · 2013-03-26 14:56:09Z

Try This

import java.io.StringReader;
import java.util.ArrayList;
import java.util.List;

import javax.xml.parsers.DocumentBuilder;
import javax.xml.parsers.DocumentBuilderFactory;

import org.w3c.dom.CharacterData;
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.InputSource;

public class XML {
  public static void main(String arg[]) throws Exception{
    String xmlRecords = "<root><x>1</x><x>2</x><x>3</x><x>4</x></root>";

    DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();
    InputSource is = new InputSource();
    is.setCharacterStream(new StringReader(xmlRecords));

    Document doc = db.parse(is);
    NodeList nodes = doc.getElementsByTagName("x");
    System.out.println(nodes.getLength());
    List<String> valueList = new ArrayList<String>();
    for (int i = 0; i < nodes.getLength(); i++) {
      Element element = (Element) nodes.item(i);

      String name = element.getTextContent();

     // Element line = (Element) name.item(0);
      System.out.println("Name: " + name);
      valueList.add(name);
    }

  }
}

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