java - 在java中无需迭代即可获取节点的xpath值

Question

我有一个 xml

<?xml version="1.0" encoding="UTF-8"?>
<information>
  <person id="1">
  <name>Deep</name>
  <age>34</age>
  <gender>Male</gender>
  </person>

 <person id="2">
  <name>Kumar</name>
  <age>24</age>
  <gender>Male</gender>
  </person>

  <person id="3">
  <name>Deepali</name>
  <age>19</age>
  <gender>Female</gender>
  </person>

  <!-- more persons... -->
</information>

DocumentBuilderFactory domFactory = DocumentBuilderFactory
                .newInstance();
        domFactory.setNamespaceAware(true);
        DocumentBuilder builder = domFactory.newDocumentBuilder();
        Document doc = builder.parse("persons.xml");
        XPath xpath = XPathFactory.newInstance().newXPath();
        // XPath Query for showing all nodes value
        XPathExpression expr = xpath.compile("//information/person[0]/name/text()");

        Object result = expr.evaluate(doc, XPathConstants.NODE);
        Node node = (Node) result;
        System.out.println(node.getNodeValue());

我需要提取我尝试上面代码的第一个人的名字，它给出了异常，任何人都可以帮助我，

已更新异常

Exception in thread "main" java.lang.ClassCastException: com.sun.org.apache.xerces.internal.dom.DeferredTextImpl cannot be cast to javax.xml.soap.Node
    at xml.main(xml.java:33)

更新答案

XPathExpression expr = xpath.compile("//information/person[1]/name");

        String str = (String) expr.evaluate(doc, XPathConstants.STRING);

        System.out.println(str);

Answer 1

XPathExpression expr = xpath.compile("//information/person[1]/name");

    String str = (String) expr.evaluate(doc, XPathConstants.STRING);

    System.out.println(str);