Java使用Saxon-HE 10.5做XSLT3转换时xsl:import的href文件找不到如何解决

问题背景 #

我正在开发一款Java应用，采用Maven项目架构引入Saxon-HE 10.5依赖，基于XSLT3实现XML转换功能。
我的主XSLT样式表通过<xsl:import>导入其他XSLT文件（示例写法：<xsl:import href="sheet1.xsl"/>），所有XSLT文件均存放于项目的./src/main/resources目录下。但程序运行时Saxon抛出FileNotFound异常，经排查发现Saxon默认在项目根目录查找导入的XSLT文件。
我希望调整Saxon查找导入文件的路径，但未找到基于s9api接口的实现方案，最初的转换代码如下：

public void transformXML(String xmlFile, String output) throws SaxonApiException, IOException, XPathExpressionException, ParserConfigurationException, SAXException {

        Processor processor = new Processor(false);
        XsltCompiler compiler = processor.newXsltCompiler();
        XsltExecutable stylesheet = compiler.compile(new StreamSource(this.getClass().getClassLoader().getResourceAsStream("transform.xsl")));

        Serializer out = processor.newSerializer(new File(output));
        out.setOutputProperty(Serializer.Property.METHOD, "text");
        Xslt30Transformer transformer = stylesheet.load30();
        transformer.transform(new StreamSource(new File(xmlFile)), out);
    }

可用解决方案 #

通过为XsltCompiler设置自定义的URIResolver，实现从类路径读取导入的XSLT资源，最终实现代码如下：

public void transformXML(String xmlFile, String output) throws SaxonApiException, IOException, XPathExpressionException, ParserConfigurationException, SAXException {

        Processor processor = new Processor(false);
        XsltCompiler compiler = processor.newXsltCompiler();
        compiler.setURIResolver(new ClasspathResourceURIResolver());
        XsltExecutable stylesheet = compiler.compile(new StreamSource(this.getClass().getClassLoader().getResourceAsStream("transform.xsl")));

        Serializer out = processor.newSerializer(new File(output));
        out.setOutputProperty(Serializer.Property.METHOD, "text");
        Xslt30Transformer transformer = stylesheet.load30();
        transformer.transform(new StreamSource(new File(xmlFile)), out);
    }
}

class ClasspathResourceURIResolver implements URIResolver
{
    @Override
    public Source resolve(String href, String base) throws TransformerException {
        return new StreamSource(this.getClass().getClassLoader().getResourceAsStream(href));
    }
}

内容的提问来源于stack exchange，提问作者DerMannMitDerMaske