如何将XSL文档中的值转换为字符串，以及如何在使用saxon解析xsl文件后删除<dot-filename>graph/node</dot-filename>标记？

<root>
    <element>Value to be converted</element>
</root>

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

    <xsl:template match="/">
        <xsl:value-of select="string(/root/element)"/>
    </xsl:template>
</xsl:stylesheet>

<root>
    <dot-filename>graph/node</dot-filename>
    <other-element>Other content</other-element>
</root>

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

    <!-- 匹配并忽略<dot-filename>节点 -->
    <xsl:template match="dot-filename"/>

    <!-- 默认模板，复制所有其他节点 -->
    <xsl:template match="@*|node()">
        <xsl:copy>
            <xsl:apply-templates select="@*|node()"/>
        </xsl:copy>
    </xsl:template>
</xsl:stylesheet>

import net.sf.saxon.s9api.*;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileOutputStream;

public class XSLTProcessor {
    public static void main(String[] args) throws Exception {
        Processor processor = new Processor(false);
        XsltCompiler compiler = processor.newXsltCompiler();
        XsltExecutable stylesheet = compiler.compile(new FileInputStream("path/to/your/stylesheet.xsl"));

        Xslt30Transformer transformer = stylesheet.load30();
        transformer.setInitialContextNode(new FileInputStream("path/to/your/input.xml").getChannel().map(FileChannel.MapMode.READ_ONLY, 0, new File("path/to/your/input.xml").length()));

        FileOutputStream outputStream = new FileOutputStream("path/to/your/output.xml");
        transformer.transform(new StreamSource(new FileInputStream("path/to/your/input.xml")), new StreamResult(outputStream));
    }
}