Hi,

Is there a technique to return the fullpath of an XML node? Ideally, I could use something similar to the treeview controls .GetFullPath method. Any guidance greatly appreciated.

My objective is to dynamically discover the lowest level in the XML structure to guide subsequent processing. The 'fullpath' approach is the best method I can think of at the moment. If anyone can provide an alternate method to 'skin the cat' I'd love to hear it.

Thanks,

Scott

The question is often asked, and different people who ask it want different things. Some people want:

a/b/c/d/e

that is, a path containing the names of the ancestor elements (easily achieved in XSLT 2.0 with string-join(ancestor-or-self::*/name(), '/')).

Some people want

a[1]/b[3]/d[4]

that is, a path that identifies an element uniquely.

And in either case, people may or may not want the result to be namespace-sensitive. This is where things get difficult: name() returns a lexical QName that cannot be used to retrieve the node again unless you know the original namespace bindings. If you want a path that can be evaluated without a namespace context, then you have two choices:

*[1]/*[3]/*[4]

or

*[namespace-uri()='http://1/' and local-name()='a'][1]/*[namespace-uri()='http://2/' and local-name()='b'][3]/....

__________________
Michael Kay
http://www.saxonica.com/
Author, XSLT 2.0 and XPath 2.0 Programmer\'s Reference