Re: XML transformation based on parameters

> I don't know how to parse the xml file inside xslt and invoke the
> transformation for each parameter...

do you want a separate output for each parameter, or one output that
has all the replacements combined?


why have this

 <xsl:template match="*[local-name()='param']">

which is likely to be slower to execute and harder to read than

 <xsl:template match="param">


Assuming you want to do all the replacements together then have a file
rep.xml that looks like

<replace>
  <r p="id" v="newvalue"/>
  <r p="id2" v="newvalue2"/>
  <r p="id3" v="newvalue3"/>
</replace>

Then 

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

 <xsl:param name="rep" select="document('rep.xml')"/>


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

 <xsl:template match="param">

  <xsl:choose>
    <xsl:when test="@name = $rep/replace/@p">
      <xsl:value-of select="$rep/replace[@p=current()/@name]/@v"/>
    </xsl:when>
    <xsl:otherwise>
      <xsl:value-of select="."/>
    </xsl:otherwise>
  </xsl:choose>

 </xsl:template>
</xsl:stylesheet>

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