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Re: XSLT: Moving node in document

From: Claudio Jolowicz <cj603@---.--.--.-->
To: NULL
Date: 4/1/2005 1:40:00 AM
Thanks, simple and elegant.

Could you give me a hint how to parameterise this?

The following approach doesn't seem to work:

------------------------------------------------------
<xsl:stylesheet
  version="1.0"
  xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

  <xsl:param name="nodeparent" />
  <xsl:param name="nodeid" />

  <xsl:template match="*">
    <xsl:copy>
      <xsl:copy-of select="@*"/>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>

  <xsl:template match="node[@id='{$nodeid}']"/>

  <xsl:template match="node[@id='{$nodeparent}']">
    <xsl:copy>
      <xsl:copy-of select="@*|//node[@id='{$nodeid}']"/>
      <xsl:apply-templates/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>
------------------------------------------------------

Easy enough to put an <xsl:choose> construct inside the identity
transform, and ignore the node satisfying "@id = $nodeid'. But how can
we then add that node to $nodeparent?

Claudio

--
  ,= ,-_-. =.    Claudio Jolowicz
 ((_/)o o(\_))   http://www.jolowicz.com
  `-'(. .)`-'
      \_/


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