XSLT - replace reoccuring parts of XML document (at any level) - xslt

Question

pros,
I need to convert the 'B' tag with 'X' tag in the following document:
<a>
<B marker="true">
<c>
<B marker="true">
<d>
<B marker="true">
</B>
<d>
</B>
</c>
</B>
</a>
Note the reoccurring 'B', it can appear at any depth in dynamic XML.
Here's what I did:
<xsl:template match="//*[#marker='true']">
<X>
<xsl:copy-of select="./node()"/>
</X>
</xsl:template>
It worked for the top-most 'B' tag, but ignored all the nested ones.
I think I know what the problem is - 'copy-of' just flushes the content of the top-most 'B' tag, without evaluating it. What can I do to make 'copy-of' reevaluate my template?
Thanks!
Baruch.

Answer 1

I would go with identity transform.
This code:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="#* | node()">
<xsl:copy>
<xsl:apply-templates select="#* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="B[#marker = 'true']">
<X>
<xsl:apply-templates/>
</X>
</xsl:template>
</xsl:stylesheet>
Against this XML input:
<a>
<B marker="true">
<c test="test">
testText
<B marker="true">
<d>
testText2
<B marker="true">
testText3
</B>
</d>
</B>
testText4
</c>
testText5
</B>
</a>
Will provide this correct result:
<a>
<X>
<c test="test">
testText
<X>
<d>
testText2
<X>
testText3
</X>
</d></X>
testText4
</c>
testText5
</X>
</a>