Re: Converting Flat XML to Hierarchical structure using xsl.

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(_dot_)pickaxe(_at_)gmail(_dot_)com> 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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