[ACCEPTED]-Passing parameters to XSLT Stylesheet via .NET-xslcompiledtransform
Accepted answer
You need to define the parameter within 7 your XSLT and you also need to pass the 6 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 5 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 4 simple XSLT sample will create just a small 3 output document containing one XML node 2 with its contents set to the value of your 1 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 3 the top of the XSLT:
<xsl:param name="Boss_ID" />
<OrgDoc>
//rest of the XSLT
</OrgDoc>
See this link
Not a 2 great example but the best I could find 1 with a quick google.
Source:
stackoverflow.com
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