如何将参数传递给XSLT？

import javax.xml.transform.*;
import javax.xml.transform.stream.StreamResult;
import javax.xml.transform.stream.StreamSource;

public class XSLTDemo {
    public static void main(String[] args) {
        try {
            TransformerFactory transformerFactory = TransformerFactory.newInstance();
            StreamSource xsltSource = new StreamSource("transform.xslt");
            Transformer transformer = transformerFactory.newTransformer(xsltSource);

            transformer.setParameter("paramName", "paramValue");

            StreamSource xmlSource = new StreamSource("input.xml");
            StreamResult outputResult = new StreamResult("output.html");
            transformer.transform(xmlSource, outputResult);
        } catch (TransformerException e) {
            e.printStackTrace();
        }
    }
}

using System.Xml;
using System.Xml.Xsl;

namespace XSLTDemo {
    class Program {
        static void Main(string[] args) {
            XslCompiledTransform transformer = new XslCompiledTransform();
            transformer.Load("transform.xslt");

            XsltArgumentList arguments = new XsltArgumentList();
            arguments.AddParam("paramName", "", "paramValue");

            transformer.Transform("input.xml", arguments, "output.html");
        }
    }
}

from lxml import etree

xslt_transformer = etree.XSLT(etree.parse("transform.xslt"))

params = {"paramName": "paramValue"}

result = xslt_transformer(etree.parse("input.xml"), **params)

with open("output.html", "wb") as f:
    f.write(etree.tostring(result, pretty_print=True))

<xsl:param name="paramName"/>