RE: Transforming node to different location in output

> Can anyone help me figure out how to transform one
> node to a different location in the output

> Example:
> Input:
> <?xml version="1.0"?>
> <document>
>   <tag1>
>     <element1>data</element1>
>     <element2>data</element2>
>   </tag1>
>   <tag2>
>     <element1>data</element1>
>     <element2>data</element2>
>   </tag2>
>   <tag3>
>     <element1>data</element1>
>     <element2>data</element2>
>   </tag3>
> </document>
> 
> Output:
> <?xml
> 
> The output that I would like to get might be something
> like this:
> 
> <?xml version="1.0" encoding="utf-8"?>
> <document>
>    <tag2>
>       <element1>data</element1>
>       <element2>data</element2>
>    </tag2>
>    <tag3>
>       <element1>data</element1>
>       <element2>data</element2>
>       <newtag1>
>          <newelement1>data</newelement1>
>          <newelement2>data</newelement2>
>       </newtag1>
>    </tag3>
> </document>
>
Hard to generalise from one example. You could achieve this particular
transformation by:
 
<xsl:template match="document">
<xsl:copy>
  <xsl:copy-of select="tag2"/>
  <xsl:copy select="tag3">
    <xsl:copy-of select="tag3/*"/>
    <xsl:apply-templates select="tag1"/>
  </xsl:copy>
</xsl:copy>
</xsl:template>

<xsl:template match="tag1">
<newtag1><xsl:apply-templates/></newtag1>
</xsl:template>

<xsl:template match="element1">
<newelement1><xsl:apply-templates/></newelement1>
</xsl:template>

general principle: xsl:copy-of for a deep copy, xsl:copy for a shallow copy,
xsl:apply-templates for a transformation.

Mike Kay


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