How to get xpath from an XmlNode instance

Question

Could someone supply some code that would get the xpath of a System.Xml.XmlNode instance?

Thanks!

Answer 1

What about using class extension ? ;) My version (building on others work) uses the syntaxe name[index]... with index omited is element has no "brothers". The loop to get the element index is outside in an independant routine (also a class extension).

Just past the following in any utility class (or in the main Program class)

static public int GetRank( this XmlNode node )
{
    // return 0 if unique, else return position 1...n in siblings with same name
    try
    {
        if( node is XmlElement ) 
        {
            int rank = 1;
            bool alone = true, found = false;

            foreach( XmlNode n in node.ParentNode.ChildNodes )
                if( n.Name == node.Name ) // sibling with same name
                {
                    if( n.Equals(node) )
                    {
                        if( ! alone ) return rank; // no need to continue
                        found = true;
                    }
                    else
                    {
                        if( found ) return rank; // no need to continue
                        alone = false;
                        rank++;
                    }
                }

        }
    }
    catch{}
    return 0;
}

static public string GetXPath( this XmlNode node )
{
    try
    {
        if( node is XmlAttribute )
            return String.Format( "{0}/@{1}", (node as XmlAttribute).OwnerElement.GetXPath(), node.Name );

        if( node is XmlText || node is XmlCDataSection )
            return node.ParentNode.GetXPath();

        if( node.ParentNode == null )   // the only node with no parent is the root node, which has no path
            return "";

        int rank = node.GetRank();
        if( rank == 0 ) return String.Format( "{0}/{1}",        node.ParentNode.GetXPath(), node.Name );
        else            return String.Format( "{0}/{1}[{2}]",   node.ParentNode.GetXPath(), node.Name, rank );
    }
    catch{}
    return "";
}   

Answer 2

If you do this, you will get a Path with Names of der Nodes AND the Position, if you have Nodes with the same name like this: "/Service[1]/System[1]/Group[1]/Folder[2]/File[2]"

public string GetXPathToNode(XmlNode node)
{         
    if (node.NodeType == XmlNodeType.Attribute)
    {             
        // attributes have an OwnerElement, not a ParentNode; also they have             
        // to be matched by name, not found by position             
        return String.Format("{0}/@{1}", GetXPathToNode(((XmlAttribute)node).OwnerElement), node.Name);
    }
    if (node.ParentNode == null)
    {             
        // the only node with no parent is the root node, which has no path
        return "";
    }

    //get the index
    int iIndex = 1;
    XmlNode xnIndex = node;
    while (xnIndex.PreviousSibling != null && xnIndex.PreviousSibling.Name == xnIndex.Name)
    {
         iIndex++;
         xnIndex = xnIndex.PreviousSibling; 
    }

    // the path to a node is the path to its parent, plus "/node()[n]", where
    // n is its position among its siblings.         
    return String.Format("{0}/{1}[{2}]", GetXPathToNode(node.ParentNode), node.Name, iIndex);
}

Answer 3

Jon's correct that there are any number of XPath expressions that will yield the same node in an an instance document. The simplest way to build an expression that unambiguously yields a specific node is a chain of node tests that use the node position in the predicate, e.g.:

/node()[0]/node()[2]/node()[6]/node()[1]/node()[2]

Obviously, this expression isn't using element names, but then if all you're trying to do is locate a node within a document, you don't need its name. It also can't be used to find attributes (because attributes aren't nodes and don't have position; you can only find them by name), but it will find all other node types.

To build this expression, you need to write a method that returns a node's position in its parent's child nodes, because XmlNode doesn't expose that as a property:

static int GetNodePosition(XmlNode child)
{
   for (int i=0; i<child.ParentNode.ChildNodes.Count; i++)
   {
       if (child.ParentNode.ChildNodes[i] == child)
       {
          // tricksy XPath, not starting its positions at 0 like a normal language
          return i + 1;
       }
   }
   throw new InvalidOperationException("Child node somehow not found in its parent's ChildNodes property.");
}

(There's probably a more elegant way to do that using LINQ, since XmlNodeList implements IEnumerable, but I'm going with what I know here.)

Then you can write a recursive method like this:

static string GetXPathToNode(XmlNode node)
{
    if (node.NodeType == XmlNodeType.Attribute)
    {
        // attributes have an OwnerElement, not a ParentNode; also they have
        // to be matched by name, not found by position
        return String.Format(
            "{0}/@{1}",
            GetXPathToNode(((XmlAttribute)node).OwnerElement),
            node.Name
            );            
    }
    if (node.ParentNode == null)
    {
        // the only node with no parent is the root node, which has no path
        return "";
    }
    // the path to a node is the path to its parent, plus "/node()[n]", where 
    // n is its position among its siblings.
    return String.Format(
        "{0}/node()[{1}]",
        GetXPathToNode(node.ParentNode),
        GetNodePosition(node)
        );
}

As you can see, I hacked in a way for it to find attributes as well.

Jon slipped in with his version while I was writing mine. There's something about his code that's going to make me rant a bit now, and I apologize in advance if it sounds like I'm ragging on Jon. (I'm not. I'm pretty sure that the list of things Jon has to learn from me is exceedingly short.) But I think the point I'm going to make is a pretty important one for anyone who works with XML to think about.

I suspect that Jon's solution emerged from something I see a lot of developers do: thinking of XML documents as trees of elements and attributes. I think this largely comes from developers whose primary use of XML is as a serialization format, because all the XML they're used to using is structured this way. You can spot these developers because they're using the terms "node" and "element" interchangeably. This leads them to come up with solutions that treat all other node types as special cases. (I was one of these guys myself for a very long time.)

This feels like it's a simplifying assumption while you're making it. But it's not. It makes problems harder and code more complex. It leads you to bypass the pieces of XML technology (like the node() function in XPath) that are specifically designed to treat all node types generically.

There's a red flag in Jon's code that would make me query it in a code review even if I didn't know what the requirements are, and that's GetElementsByTagName. Whenever I see that method in use, the question that leaps to mind is always "why does it have to be an element?" And the answer is very often "oh, does this code need to handle text nodes too?"

Answer 4

There's no such thing as "the" xpath of a node. For any given node there may well be many xpath expressions which will match it.

You can probably work up the tree to build up an expression which will match it, taking into account the index of particular elements etc, but it's not going to be terribly nice code.

Why do you need this? There may be a better solution.

Answer 5

I had to do this recently. Only elements needed to be considered. This is what I came up with:

    private string GetPath(XmlElement el)
    {
        List<string> pathList = new List<string>();
        XmlNode node = el;
        while (node is XmlElement)
        {
            pathList.Add(node.Name);
            node = node.ParentNode;
        }
        pathList.Reverse();
        string[] nodeNames = pathList.ToArray();
        return String.Join("/", nodeNames);
    }

Answer 6

My 10p worth is a hybrid of Robert and Corey's answers. I can only claim credit for the actual typing of the extra lines of code.

    private static string GetXPathToNode(XmlNode node)
    {
        if (node.NodeType == XmlNodeType.Attribute)
        {
            // attributes have an OwnerElement, not a ParentNode; also they have
            // to be matched by name, not found by position
            return String.Format(
                "{0}/@{1}",
                GetXPathToNode(((XmlAttribute)node).OwnerElement),
                node.Name
                );
        }
        if (node.ParentNode == null)
        {
            // the only node with no parent is the root node, which has no path
            return "";
        }
        //get the index
        int iIndex = 1;
        XmlNode xnIndex = node;
        while (xnIndex.PreviousSibling != null) { iIndex++; xnIndex = xnIndex.PreviousSibling; }
        // the path to a node is the path to its parent, plus "/node()[n]", where 
        // n is its position among its siblings.
        return String.Format(
            "{0}/node()[{1}]",
            GetXPathToNode(node.ParentNode),
            iIndex
            );
    }