Obtain Full Path of XML Node

[connah]

Hi All!

How can I obtain the full XML path of any given XML Node? For example,
if my XML document looks like this:

---BEGIN XML DOC---
<Tag1>
<Tag2>
<Tag3></Tag3>
</Tag2>
</Tag1>
---END XML DOC---

And my code looks like this:

---BEGIN VB6 CODE---
Dim objCustNode As MSXML2.IXMLDOMNode
Dim objCustNodeList As MSXML2.IXMLDOMNodeList

objCustXML.Load("myFile.xml")
Set objCustNodeList = objCustXML.selectNodes("//*")
For Each objCustNode In objCustNodeList
Debug.Print objCustNode.nodeName
Next
---END VB6 CODE---

At the point in the For Next loop where objCustNode.nodeName is equal
to "Tag3", I want to know what's it's FULL path is, ie,
"//Tag1/Tag2/Tag3".

How can I do this? Thank you!

[Joe Kesselman]

> At the point in the For Next loop where objCustNode.nodeName is equal

to "Tag3", I want to know what's it's FULL path is, ie,
"//Tag1/Tag2/Tag3".

If you want absolute path, you want to start with /, not //

It's fairly trivial to implement a non-namespace-aware version of this,
walking up the node's ancestors and building up the path. Note that at
each level you need to count the number of preceeding instances of nodes
with the same type and name, since what you really want is more like
/Tag1[1]/Tag2[4]/Tag3[2]

Properly supporting namespaces, unfortunately, is uglier, since the
normal XPath syntax requires that the namespace bindings be given as
prefixes but provides no syntax for declaring those prefixes. The ugly
workaround is to generate something more like
/child::*[localname()="Tag1" and namespace()="http://whatever"][1]
for each step.

An XSLT implementation of the non-namespace-aware version appeared in my
DeveloperWorks article on using stylesheets to debug other stylesheets.
It shouldn't be hard to implement it in any other language, and
off-the-shelf implementations probably do exist though I've had no
reason to go looking for one recently. I don't use VB so I can't give
you any specific advice for that environment.

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