Extract node value by calculating absolute position - xslt

Question

My source XML looks:
<test>
<text1>Test</text1>
<text2>Test</text2>
<text2>Test</text2>
<section>
<text1>Test<bold>content</bold></text1>
<text1>Test</text1>
<text2>Test</text2>
<text2>Test</text2>
</section>
</test>
Want to extract the value of 6th node, based on the absolute number of the element (overall count). The absolute number of the element has been identified using <xsl:number level="any" from="/" count="*"/>.

Answer 1

The XPath expression /descendant::*[6] should give you the element you need.
<xsl:template match="/">
<xsl:copy-of select="/descendant::*[6]" />
</xsl:template>
outputs
<text1>Test<bold>content</bold></text1>
Note that this is an example of the difference between descendant:: and // - //*[6] would give you all elements that are the sixth child element of their respective parent, rather than simply the sixth element in the document in depth-first order.

Answer 2

This xslt
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xs="http://www.w3.org/2001/XMLSchema" xmlns:fn="http://www.w3.org/2005/xpath-functions">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:variable name="allElements" select="//element()" />
<xsl:template match="/">
<output>
<xsl:value-of select="$allElements[6]" />
</output>
</xsl:template>
</xsl:stylesheet>
will result in
<?xml version="1.0" encoding="UTF-8"?>
<output>Testcontent</output>