Passing parameters to XSLT Stylesheet via .NET

懵懂的女人 提交于 2019-11-27 01:07:57

You need to define the parameter within your XSLT and you also need to pass the XsltArgumentList as an argument to the Transform call:

private static void CreateHierarchy(string manID)
{
    string man_ID = manID;

    XsltArgumentList argsList = new XsltArgumentList();
    argsList.AddParam("Boss_ID", "", man_ID);

    XslCompiledTransform transform = new XslCompiledTransform(true);
    transform.Load("htransform.xslt");

    using (StreamWriter sw = new StreamWriter("output.xml"))
    {
        transform.Transform("LU AIB.xml", argsList, sw);
    }
}

Please note that the xsl:param must be defined below the xsl:stylesheet element:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="xml" indent="yes" />

  <xsl:param name="Boss_ID"></xsl:param>

  <xsl:template match="OrgDoc">

     <!-- template body goes here -->

  </xsl:template>


</xsl:stylesheet>

This simple XSLT sample will create just a small output document containing one XML node with its contents set to the value of your parameter. Have a try:

<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
  <xsl:output method="xml" indent="yes" />
  <xsl:param name="Boss_ID"></xsl:param>

  <xsl:template match="/">
    <out>
      <xsl:value-of select="$Boss_ID" />
    </out>
  </xsl:template>

</xsl:stylesheet>

you probably need to define the param at the top of the XSLT:

<xsl:param name="Boss_ID" />
<OrgDoc>
 //rest of the XSLT
</OrgDoc>

See this link

http://projects.ischool.washington.edu/tabrooks/545/2004Autumn/ContentManagement/PassingParameters.htm

Not a great example but the best I could find with a quick google.

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