JAVA - How can i read an XML file and save each node to a string array

Here is a simple code that can achieve what do you wanted. I used Lists in this code. To get just data you wanted, i do not have any other method than checking the nodes than have been read from the xml file.

 import java.io.File;
    import java.io.IOException;
    import java.util.ArrayList;
    import java.util.List;
    import org.jdom.Document;
    import org.jdom.Element;
    import org.jdom.JDOMException;
    import org.jdom.input.SAXBuilder;



    public class readConfigXML {

 // the nodes that we want to be read
        public static List<String>  wantedNodes  = new ArrayList<String>() ;
// the attributes that we want to be read
        public static List<String>  wantedAttributes= new ArrayList<String>()   ;
        // init the nodes and attributes wanted here
        public static void init(){
            wantedNodes.add("HomeScreenName") ;
            wantedNodes.add("HomeScreenAttributes") ;
            wantedAttributes.add("Logo") ; // ...
        }

        public static void main(String[] args) 
        {
            // here init your wanted nodes and attributes
            init() ;

          SAXBuilder builder = new SAXBuilder();
          String folderPath = "C:\\Users\\udi\\Documents\\external\\XML\\";
          String fileName = "configTest.xml";
          String filePath = folderPath + fileName;  
          File xmlFile = new File(filePath);

          List<String> nodes = new ArrayList<String>() ;
          List<String> attributeNodes = new ArrayList<String>() ;

          try {
                Document document = (Document) builder.build(xmlFile);
                Element rootNode = document.getRootElement();
                List configList = rootNode.getChildren();
                for (int i = 0; i < configList.size(); i++) 
                {
                    Element node = (Element) configList.get(i);
                    // check if this node is wanted  
                    if(wantedNodes.contains(node.getName())){
                        nodes.add(node.getName()) ;
                        List dataNodes = node.getChildren();
                        for (int j = 0; j < dataNodes.size(); ++j) 
                        {
                            Element dataNode = (Element) dataNodes.get(j);
                            // check if this attribute is wanted 
                            if(wantedAttributes.contains(dataNode.getName())){
                                attributeNodes.add(dataNode.getValue()) ;
                            }
                        }
                    }
                }  

            }
          catch (IOException io) 
          {
            System.out.println(io.getMessage());
          } 
          catch (JDOMException jdomex) 
          {
            System.out.println(jdomex.getMessage());
          }
        }

    }

for the standard JDK package org.w3c.dom(I did not check but perhaps your org.jdom must have the same or similar:

1. How to get just the data i want and not everything from the XML.

each node has a node type. Then it may have a child of TEXT_NODE type. (Node Type is short and Node interface has constants for types. e.g.

if (dataNode.getNodeType() == Node.TEXT_NODE)

So when you traverse through node children - check if it is a TEXT_NODE then it is a data you need. e.g. Element(node) <Logo>Logo</Logo> has a child text node with value "Logo" and you can get it buy calling: node.getTextContent().

BTW: Be careful with getTextContent() if it is mixed node (with text and other child elements) it will return texts from all children. I do not see it in your example, but first check the Node Type if it is Text Node you are good to use it.

2. How to save the data (I wanted arrays, but i'm open to suggestions).

It is up to you, but as long as Java does not have dynamic arrays it is better to use List as example LinkedList or ArrayList to capture data from unknown numbers of elements.

If you still want to have array, after all done get array from your resulting List by calling its toArray() method.

Also, for general purpose (if XML tree has unknown or large number of levels you need to use recursion to process child nodes). If it is not a case and you know exactly how deep your XML is, you can be fine to use nested for loops, but still recursion is preferred.

So the best way i could find was to edit my XML so that all the main child nodes will have the same name with different attributes like this:

<Page id="Home">
<Page id="Sign in">

And the code to get all the main Nodes names is this:

public static void main(String[] args) 
{
    String folderPath = "C:\\XML\\";
    String fileName = "2.xml";
    String xmlFile = folderPath + fileName;         
    DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
    DocumentBuilder dBuilder;
    try
    /// gets the main nodes in the config and saves them to a list
    {
        dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(xmlFile);
        doc.getDocumentElement().normalize();
        //System.out.println("Root element is:" + doc.getDocumentElement().getNodeName());
        NodeList mainList = doc.getElementsByTagName("Page");
        List<String> list = new ArrayList<String>();
        for (int temp = 0; temp < mainList.getLength(); temp++)
        {
            Node mainNode = mainList.item(temp);
            Element eElement = (Element) mainNode;
            list.add(eElement.getAttribute("id"));
            System.out.println( list.get(temp));
        }
        System.out.println(list);
        System.out.println("----------------------------------------");

    }
    catch(SAXException | ParserConfigurationException | IOException e1)
    {
        e1.printStackTrace();
    }

}

And the output will be:

Home Sign in Register Login and as a list: [Home, Sign in, Register, Login]