如何在Groovy或Java中按“路径”元素过滤XML字符串

import groovy.util.XmlSlurper

def xmlString = '''
<root>
  <element1>
    <subelement>Value 1</subelement>
  </element1>
  <element2>
    <subelement>Value 2</subelement>
  </element2>
</root>
'''

def xml = new XmlSlurper().parseText(xmlString)
def filteredElements = xml.'**'.findAll { node ->
  node.name() == 'subelement'
}

filteredElements.each { element ->
  println element.text()
}

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.xpath.*;

import org.w3c.dom.Document;
import org.w3c.dom.NodeList;

public class XPathExample {
    public static void main(String[] args) throws Exception {
        String xmlString = "<root><element1><subelement>Value 1</subelement></element1><element2><subelement>Value 2</subelement></element2></root>";

        DocumentBuilderFactory factory = DocumentBuilderFactory.newInstance();
        Document document = factory.newDocumentBuilder().parse(new InputSource(new StringReader(xmlString)));

        XPathFactory xPathFactory = XPathFactory.newInstance();
        XPath xPath = xPathFactory.newXPath();
        XPathExpression expression = xPath.compile("//subelement");

        NodeList nodeList = (NodeList) expression.evaluate(document, XPathConstants.NODESET);
        for (int i = 0; i < nodeList.getLength(); i++) {
            System.out.println(nodeList.item(i).getTextContent());
        }
    }
}