Read the nodes that have the same value as the subnodes - XML

It is more of a general JAVA / XML problem, but given that it is going into my BlackBerry app I thought I'd see if anyone knows.

Consider a simple XML document:

                    Whatever 1                   Whatever 2               Whatever 3

Using the standard org.w3c.dom, I can get the nodes in X by the practice...

NodeList fullnodelist = doc.getElementsByTagName ("x");

But if I want to go to the next set of 'e', I try to use something like...

Element element = (Element) fullnodelist.item(0);NodeList nodes = pelement.getElementsByTagName("e");

EXPECTED back '3' nodes (because there are 3 series of 'e'), but it returns '9' - because it gets all entries including the 'e' apperently.

It would be nice in the above case, because I could probably go through and find what I'm looking for. The problem I have is that when the XML file looks like the following:

            whatever              Something Else                    whatever              Something Else

When I ask 'e' value, it returns 4, instead of (what I want) 2.

I am simply not understand how DOM parsing works? Generally, in the past I used my own XML documents so I name never articles like this, but unfortunately this isn't my XML file and I don't have the choice to work like this.

What I thought I would do, it is write a loop knots "drills down" so that I can combine each node...

  public static NodeList getNodeList(Element pelement, String find)
    {
        String[] nodesfind = Utilities.Split(find, "/");
        NodeList nodeList = null;

        for (int i = 0 ; i <= nodesfind.length - 1; i++ )
        {
            nodeList = pelement.getElementsByTagName( nodesfind[i] );
            pelement = (Element)nodeList.item(i);
        }

        // value of the nod we are looking for
        return nodeList;
    }

.. While if adopted you ' s/e' in the service, he would return the 2 nodes I'm looking (or elements, perhaps I'm using the wrong terminology?). on the contrary, it returns all the 'e' nodes in this node.

Anyway, if anyone is still with me and has a suggestion, it would be appreciated.

Well, there is no doubt that there is a learning curve robust for XML programming. You can take an hour or two and go through one of the tutorials that are circulating on the net. (Like that of w3schools.com.)

Basically, almost everything in XML is a node, the Document that returns the parser. The API for node tells you that you can test the node type you have by calling getNodeType, which returns one of the constants of type node (Node.ELEMENT_NODE, Node.TEXT_NODE, etc..) If necessary, you can then convert the variable to the corresponding interface (element, text, etc.).

Similarly, the API documentation say you for any node, calling getChildNodes (or for an element node or Document getElementsByTagName) will give you a NodeList (a little non-types of nodes in the XML API), while calling getFirstChild and getNextSibling to any node will give you another node (or null ).

Once you learn the API, writing logic of course is not all that hard. For example, if the only 'e' interest tags are those directly under the element root of the document (as shown in your example) you can simply go to them directly:

Vector getTopENodes(Document doc) {  Vector vec = new Vector();  NodeList nodes = doc.getDocumentElement().getChildNodes();  int n = nodes.getLength();  for (int i = 0; i < n; ++i) {    Node node = nodes.item(i);    if (node.getNodeType() == Node.ELEMENT_NODE &&        "e".equals(node.getNodeName()))    {      vec.addElement(node);    }  }  return vec;}

Note that this example does not assume that all children are nodes of element 'e '. the document could have comments, white space or something else that makes it into the DOM as comment, text or any other type of node.

On the other hand, if you want to capture every "e" tag which is directly under the ' tag, no matter the level, then you need to do something a little more complicated (it's on the top of my head - no guarantee):

static class NodeListImp implements NodeList {  private Vector nodes = new Vector();  public int getLength() {    return nodes.size();  }  public Node item(int index) {    return (Node) nodes.elementAt(index);  }  public add(Node node) {    nodes.addElement(node);  }}

NodeList getTargetNodes(Document doc) {  NodeListImp list = new NodeListImp();  getTargetnodes(list, doc.getDocumentElement(), false);  return list;}

void getTargetNodes(NodeListImp list, Node node, boolean parentIsS) {  if (node.getNodeType() == Node.ELEMENT_NODE) {    // node name is tag name for element nodes    String name = node.getNodeName();    if (parentIsS && "e".equals(name)) {      list.add(node);    }    parentIsS = "s".equals(name);    for (Node child = node.getFirstChild();         child != null;         child = child.getNextSibling())    {      getTargetNodes(list, child, parentIsS);    }  }}

I hope that it gets the idea across.

Tags: BlackBerry Developers

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