RE: xslt: Wrapping multiple elements in a parent element

Steve,

I have solved this puzzle.

I have the following xml file :

<?xml version="1.0"?>
<Root>
        <a>
                <b>
                        <c>
                        <d>sundar</d>
                        <e>
                                <f>fff</f>
                        </e>
                        </c>
                        <d>sundar2</d>
                </b>
        </a>
        <a>
                <b>
                        <d>sundar3</d>
                </b>
        </a>
        <d>sundar4</d>                  
</Root>

I am trying to group all the <d> elements separately and rest of the
document will be same.
<d> elements will be grouped under <D> element.

This is the stylesheet which does that:

<?xml version="1.0"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
  <xsl:output method="html" indent="yes" />     
  
<xsl:template match="/Root">
<Root>
        <xsl:apply-templates select="*[name(.)!='d']"/>
        <D>
        <xsl:apply-templates select="//d"/>
        </D>
</Root>
</xsl:template>

<xsl:template match="*">
        <xsl:element name="{name(.)}">
                <xsl:value-of select="text()"/>
        <xsl:apply-templates select="*[name(.)!='d']"/>
        </xsl:element>
</xsl:template>

<xsl:template match="d">
        <d>
                <xsl:value-of select="text()"/>
        </d>
</xsl:template>
        
</xsl:stylesheet>

Hope this helps you.

sundar

-----Original Message-----
From: Steven Curry [mailto:scurry1(_at_)msn(_dot_)com]
Sent: Friday, May 09, 2003 7:44 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] xslt: Wrapping multiple elements in a parent element


I need to do an xml to xml transformation where I take 1 or more elements of

a certain name (they are grouped together somewhere in the hierarchy) and 
wrap them in a new parent element.  The rest of the document should stay the

same.  Any xslt suggestions?  I'm sure there is any easy solution but I 
haven't found it yet.

Thanks,

Steve

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