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At 2011-01-08 17:20 +0530, pankaj.c@xxxxxxxxxxxxxxxxxx wrote:
Is there a way if I have XML. You don't say how your name= attributes are derived, so I guessed. The answer below using XSLT 2 is quite straightforward. I hope this helps. . . . . . . . . . . Ken T:\ftemp>type pankaj.xml
<my_group count="3"/>
T:\ftemp>type pankaj.xsl
<?xml version="1.0" encoding="US-ASCII"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xsd="http://www.w3.org/2001/XMLSchema"
exclude-result-prefixes="xsd"
version="2.0"><xsl:output indent="yes"/> <xsl:template match="my_group">
<xsl:copy>
<xsl:copy-of select="@*"/>
<xsl:for-each select="1 to xsd:integer(@count)">
<group name="type{.}"/>
</xsl:for-each>
</xsl:copy>
</xsl:template></xsl:stylesheet> T:\ftemp>xslt2 pankaj.xml pankaj.xsl <?xml version="1.0" encoding="UTF-8"?> <my_group count="3"> <group name="type1"/> <group name="type2"/> <group name="type3"/> </my_group> T:\ftemp>
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