Get XML only immediate children elements by name

I realise you found something of a solution to this in May @kentcdodds but I just had a fairly similar problem which I've now found, I think (perhaps in my usecase, but not in yours), a solution to.

a very simplistic example of my XML format is shown below:-

<?xml version="1.0" encoding="utf-8"?><rels>    <relationship num="1">        <relationship num="2">            <relationship num="2.1"/>            <relationship num="2.2"/>        </relationship>    </relationship>    <relationship num="1.1"/>    <relationship num="1.2"/></rels>

As you can hopefully see from this snippet, the format I want can have N-levels of nesting for [relationship] nodes, so obviously the problem I had with Node.getChildNodes() was that I was getting all nodes from all levels of the hierarchy, and without any sort of hint as to Node depth.

Looking at the API for a while , I noticed there are actually two other methods that might be of some use:-

• Node.getFirstChild()
• Node.getNextSibling()

Together, these two methods seemed to offer everything that was required to get all of the immediate descendant elements of a Node. The following jsp code should give a fairly basic idea of how to implement this. Sorry for the JSP. I'm rolling this into a bean now but didn't have time to create a fully working version from picked apart code.

<%@page import="javax.xml.parsers.DocumentBuilderFactory,                javax.xml.parsers.DocumentBuilder,                org.w3c.dom.Document,                org.w3c.dom.NodeList,                org.w3c.dom.Node,                org.w3c.dom.Element,                java.io.File" %><% try {    File fXmlFile = new File(application.getRealPath("/") + "/utils/forms-testbench/dom-test/test.xml");    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();    DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();    Document doc = dBuilder.parse(fXmlFile);    doc.getDocumentElement().normalize();    Element docEl = doc.getDocumentElement();           Node childNode = docEl.getFirstChild();         while( childNode.getNextSibling()!=null ){                  childNode = childNode.getNextSibling();                 if (childNode.getNodeType() == Node.ELEMENT_NODE) {                     Element childElement = (Element) childNode;                         out.println("NODE num:-" + childElement.getAttribute("num") + "<br/>\n" );                  }           }} catch (Exception e) {    out.println("ERROR:- " + e.toString() + "<br/>\n");}%>

This code would give the following output, showing only direct child elements of the initial root node.

NODE num:-1NODE num:-1.1NODE num:-1.2

Hope this helps someone anyway. Cheers for the initial post.

You can use XPath for this, using two path to get them and process them differently.

To get the <file> nodes direct children of <notification> use //notification/file and for the ones in <group> use //groups/group/file.

This is a simple sample:

public class SO10689900 {    public static void main(String[] args) throws Exception {        DocumentBuilder db = DocumentBuilderFactory.newInstance().newDocumentBuilder();        Document doc = db.parse(new InputSource(new StringReader("<notifications>\n" +                 "  <notification>\n" +                 "    <groups>\n" +                 "      <group name=\"zip-group.zip\" zip=\"true\">\n" +                 "        <file location=\"C:\\valid\\directory\\\" />\n" +                 "        <file location=\"C:\\this\\file\\doesn't\\exist.grr\" />\n" +                 "        <file location=\"C:\\valid\\file\\here.txt\" />\n" +                 "      </group>\n" +                 "    </groups>\n" +                 "    <file location=\"C:\\valid\\file.txt\" />\n" +                 "    <file location=\"C:\\valid\\file.xml\" />\n" +                 "    <file location=\"C:\\valid\\file.doc\" />\n" +                 "  </notification>\n" +                 "</notifications>")));        XPath xpath = XPathFactory.newInstance().newXPath();        XPathExpression expr1 = xpath.compile("//notification/file");        NodeList nodes = (NodeList)expr1.evaluate(doc, XPathConstants.NODESET);        System.out.println("Files in //notification");        printFiles(nodes);        XPathExpression expr2 = xpath.compile("//groups/group/file");        NodeList nodes2 = (NodeList)expr2.evaluate(doc, XPathConstants.NODESET);        System.out.println("Files in //groups/group");        printFiles(nodes2);    }    public static void printFiles(NodeList nodes) {        for (int i = 0; i < nodes.getLength(); ++i) {            Node file = nodes.item(i);            System.out.println(file.getAttributes().getNamedItem("location"));        }    }}

It should output:

Files in //notificationlocation="C:\valid\file.txt"location="C:\valid\file.xml"location="C:\valid\file.doc"Files in //groups/grouplocation="C:\valid\directory\"location="C:\this\file\doesn't\exist.grr"location="C:\valid\file\here.txt"

Well, the DOM solution to this question is actually pretty simple, even if it's not too elegant.

When I iterate through the filesNodeList, which is returned when I call notificationElement.getElementsByTagName("file"), I just check whether the parent node's name is "notification". If it isn't then I ignore it because that will be handled by the <group> element. Here's my code solution:

for (int j = 0; j < filesNodeList.getLength(); j++) {  Element fileElement = (Element) filesNodeList.item(j);  if (!fileElement.getParentNode().getNodeName().equals("notification")) {    continue;  }  ...}