Re: Converting Flat XML to Hierarchical structure usin

Thanks for the suggestion..

I want to get the output in Hierarchical order using
the text().

I have modified the XSL, it follows:
<?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"/>
<xsl:template match="office">
<layout>
 <xsl:apply-templates
select="ordered-list/list-item"/>
</layout>				
</xsl:template>

<xsl:template match="list-item">
 <xsl:choose>
   <xsl:when test="./h">
       <xsl:element name="{./h}">
	 <xsl:apply-templates
select="ordered-list/list-item"/>
	</xsl:element>
   </xsl:when>
   <xsl:otherwise>
	<xsl:apply-templates
select="ordered-list/list-item"/>
   </xsl:otherwise>
  </xsl:choose>	
</xsl:template>
</xsl:stylesheet>

I got the following XML:
<?xml version="1.0" encoding="UTF-8"?>
<layout>
	<Application/>
	<Processgroup/>
	<Process/>
	<PanelName/>
</layout>

But my expected output is:
<layout>
	<Application/>
	  <Processgroup>
	    <Process>
  	      <PanelName/>
	    </Process>
	 </Processgroup>
	</Application>
</layout>

Regards,
Raj
--- Ragulf Pickaxe <ragulf.pickaxe@xxxxxxxxx> wrote:

> Raj,
> 
> >    <xsl:apply-templates
> select="ordered-list/list-item/"/>
> 
> If you want to select list-item elements, the above
> should be:
> <xsl:apply-templates
> select="ordered-list/list-item"/> (that is,
> without the trailing /).
> 
> From the output I expect that you get the default
> templates. That
> would then be why you get the text() elements in the
> output.
> 
> The match="ordered-list/list-item/" should be
> match="ordered-list/list-item".
> And in this example, you could as well have
> match="list-item".
> 
> Ragulf Pickaxe :-)
> 
> 



		
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