[xsl] XPath to find duplicate elements

["Leo Studer leo.studer@xxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Hello

I have an xml file where I need and XPath expression to find all elements that
are a copy of a previous element in the same file.

I came up with the following:


let $xml := <a>
                <b>
                    <c>1</c>
                    <d>1</d>
                    <d>2</d>
                </b>
                <b>
                    <c>2</c>
                    <d>1</d>
                    <d>3</d>
                </b>
                <b>
                    <c>3</c>
                    <d>2</d>
                    <d>3</d>
                </b>
                <b>
                    <c>3</c>
                    <d>2</d>
                    <d>3</d>
                </b>
            </a>
            return

$xml//*[let $node1 := . return some $node2 in $xml//*[. << $node1] satisfies
deep-equal($node1,$node2)]/(ancestor-or-self::node()/concat(node-name(.),1+co
unt(preceding-sibling::*)),'&#10;')

which gives

 a1 b2 d2
 a1 b3 d2
 a1 b3 d3
 a1 b4
 a1 b4 c1
 a1 b4 d2
 a1 b4 d3


This is actually not to bad. However, I would like the correct position of the
copy. Instead of preceding-sibling::* in the count() function, I would like to
write something like preceding-sibling::element(node-name(.)). Unfortunately
the processor does not allow that ;-(b&

Any suggestions?

Thanks in advance
Leo