Get XPath of XML Tag

Question

If I have an XML document like below:

<foo>
   <foo1>Foo Test 1</foo1>
   <foo2>
       <another1>
           <test10>This is a duplicate</test10>
       </another1>
   </foo2>
   <foo2>
       <another1>
           <test1>Foo Test 2</test1>
       </another1>
   </foo2>
   <foo3>Foo Test 3</foo3>
   <foo4>Foo Test 4</foo4>
</foo>

How do I get the XPath of <test1> for example? So the output should be something like: foo/foo2[2]/another1/test1

I'm guessing the code would look something like this:

public String getXPath(Document document, String xmlTag) {
   String xpath = "";
   ...
   //Get the node from xmlTag
   //Get the xpath using the node
   return xpath;
}

Let's say String XPathVar = getXPath(document, "<test1>"); . I need to get back an absolute xpath that will work in the following code:

XPath xpath = XPathFactory.newInstance().newXPath(); 
XPathExpression xpr = xpath.compile(XPathVar); 
xpr.evaluate(Document, XPathConstants.STRING); 

But it can't be a shortcut like //test1 because it will also be used for meta data purposes.

When printing the result out via:

System.out.println(xpr.evaluate(Document, XPathConstants.STRING));

I should get the node's value. So if XPathVar = foo/foo2[2]/another1/test1 then I should get back:

Foo Test 2 and not This is a duplicate

Answer 1

You don't 'get' an xpath in the same way you don't 'get' sql.

An xpath is a query you write based on your understanding of an xml document or schema, just as sql is a query you write based on your understanding of a database schema - you don't 'get' either of them.

I would be possible to generate xpath statements from the DOM simply by walking back up the nodes from a given node, though to do this generically enough, taking into account attribute values on each node, would make the resulting code next to useless. For example (which comes with a warning that this will find the first node that has a given name, xpath is much more that this and you may as well just use the xpath //foo2 ):

import java.io.ByteArrayInputStream;
import javax.xml.parsers.DocumentBuilderFactory;

import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;

public class XPathExample 
{
  private static String getXPath(Node root, String elementName)
  {
    for (int i = 0; i < root.getChildNodes().getLength(); i++)
    {
      Node node = root.getChildNodes().item(i);

      if (node instanceof Element)
      {
        if (node.getNodeName().equals(elementName))
        {
          return "/" + node.getNodeName();
        }
        else if (node.getChildNodes().getLength() > 0)
        {
          String xpath = getXPath(node, elementName);
          if (xpath != null)
          {
            return "/" + node.getNodeName() + xpath;
          }
        }
      }
    }

    return null;
  }

  private static String getXPath(Document document, String elementName)
  {
    return document.getDocumentElement().getNodeName() + getXPath(document.getDocumentElement(), elementName);
  }

  public static void main(String[] args) 
  {
    try 
    {
      Document document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(
        new ByteArrayInputStream(
          ("<foo><foo1>Foo Test 1</foo1><foo2><another1><test1>Foo Test 2</test1></another1></foo2><foo3>Foo Test 3</foo3><foo4>Foo Test 4</foo4></foo>").getBytes()
        )
      );

      String xpath = "/" + getXPath(document, "test1");
      System.out.println(xpath);        

      Node node1 = (Node)XPathFactory.newInstance().newXPath().compile(xpath).evaluate(document, XPathConstants.NODE);
      Node node2 = (Node)XPathFactory.newInstance().newXPath().compile("//test1").evaluate(document, XPathConstants.NODE);

      //This evaluates to true, hence you may as well just use the xpath //test1.
      System.out.println(node1.equals(node2));
    }
    catch (Exception e) 
    {
      e.printStackTrace();
    }
  }
}

Likewise you could write an XML transformation that turned an xml document into a series of xpath statements but this transformation would be more complicated that writing the xpath in the first place and so largely pointless.

Answer 2

How's this:

private static String getXPath(Document root, String elementName)
{
  try{
      XPathExpression expr = XPathFactory.newInstance().newXPath().compile("//" + elementName);
      Node node = (Node)expr.evaluate(root, XPathConstants.NODE);

      if(node != null) {
          return getXPath(node);
      }
  }
  catch(XPathExpressionException e) { }

  return null;
}

private static String getXPath(Node node) {
    if(node == null || node.getNodeType() != Node.ELEMENT_NODE) {
        return "";
    }

    return getXPath(node.getParentNode()) + "/" + node.getNodeName();
}

Note that this is first locating the node (using XPath) and then using the located node to get its XPath. Quite the roundabout approach to get a value you already have.

Working ideone example: http://ideone.com/EL4783