Re: [xsl] reversing or swapping nested node hierarchy

["G. Ken Holman" <gkholman@xxxxxxxxxxxxxxxxxxxx>]

At 2006-03-19 13:40 -0500, Chris Capon wrote:
> In an XML document:
> ...
> Is it possible to reverse the a/b hierarchy to produce b/a on output?
> To complicate the problem, not all <a>'s have <b>'s.
>
> So, in solving the problem, I've come up with something like this:
> ...
> ..which works, but the <b> template now has working knowledge of <a>.
>
> I'd like to find a generic solution where <b> doesn't have to know about
> <a> so that <b> can be nested inside other elements too and yet work the
> same way.

I hope the code below helps.  You don't say what you want done with 
attributes, so I didn't do anything with them myself.

. . . . . . . . . Ken

T:\ftemp>type capon.xml
<root>
  <a>
    <b />
  </a>
  <a />
  <d>
    <b />
  </d>
  <d/>
  <c>
    <b />
  </c>
</root>

T:\ftemp>xslt capon.xml capon.xsl con
<?xml version="1.0" encoding="utf-8"?><root>
  <b><a/></b>
  <a/>
  <b><d/></b>
  <d/>
  <b><c/></b>
</root>
T:\ftemp>type capon.xsl
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
                version="1.0">

<xsl:template match="@*|node()"><!--identity for all other nodes-->
  <xsl:copy>
    <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
</xsl:template>

<xsl:template match="*[b]">
  <b>
    <xsl:copy/>
  </b>
</xsl:template>

</xsl:stylesheet>
T:\ftemp>

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