How to read XML using XPath in Java

Question

I want to read XML data using XPath in Java, so for the information I have gathered I am not able to parse XML according to my requirement.

here is what I want to do

Answer 1

If you have a xml like below

    
        
            
                Testabc
                12121
                Testpqr
            
        
    
    
        
            12--abc--pqr
        
    

and wanted to extract the below xml

   
      
         Testabc
         12121
         Testpqr
      
   

The below code helps to achieve the same

public static void main(String[] args) {

    File fXmlFile = new File("C://Users//abhijitb//Desktop//Test.xml");
    DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
    Document document;
    Node result = null;
    try {
        document = dbf.newDocumentBuilder().parse(fXmlFile);
        XPath xPath = XPathFactory.newInstance().newXPath();
        String xpathStr = "//Envelope//Header";
        result = (Node) xPath.evaluate(xpathStr, document, XPathConstants.NODE);
        System.out.println(nodeToString(result));
    } catch (SAXException | IOException | ParserConfigurationException | XPathExpressionException
            | TransformerException e) {
        e.printStackTrace();
    }
}

private static String nodeToString(Node node) throws TransformerException {
    StringWriter buf = new StringWriter();
    Transformer xform = TransformerFactory.newInstance().newTransformer();
    xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
    xform.transform(new DOMSource(node), new StreamResult(buf));
    return (buf.toString());
}

Now if you want only the xml like below

   
      Testabc
      12121
      Testpqr
   

You need to change the

String xpathStr = "//Envelope//Header"; to String xpathStr = "//Envelope//Header/*";