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="https://stackoverflow.com/">
<out>
<xsl:value-of select="$Boss_ID" />
</out>
</xsl:template>
</xsl:stylesheet>