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Re: [xsl] generating xml from xslt

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Sarkup Sarkup wrote:

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

Define three global parameters
<xsl:param name="a1"/>
<xsl:param name="a2"/>
<xsl:param name="a3"/>
so that you can provide values for the transformation then keep this template

 <xsl:template match="@*|node()">
  <xsl:copy>
   <xsl:apply-templates select="@*|node()"/>
  </xsl:copy>
 </xsl:template>

remove the following template

<xsl:template match="/root/status/locn/loc1/address">
<xsl:element name="add1"></xsl:element>
<xsl:element name="add2"></xsl:element>
<xsl:element name="add3"></xsl:element>
</xsl:template>

and add three templates
  <xsl:template match="add1">
    <xsl:copy>
      <xsl:value-of select="$a1"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="add2">
    <xsl:copy>
      <xsl:value-of select="$a2"/>
    </xsl:copy>
  </xsl:template>
  <xsl:template match="add3">
    <xsl:copy>
      <xsl:value-of select="$a3"/>
    </xsl:copy>
  </xsl:template>
--

	Martin Honnen
	http://JavaScript.FAQTs.com/

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