我在android中解析XML时遇到异常

代码是

    ///  Element rootelement = doc.getDocumentElement();
    XPathFactory xpathfactory = XPathFactory.newInstance();
    XPath xpath = xpathfactory.newXPath();
    try {
         xpathexpression = xpath.compile("//@*[name()='diffgr:id']");
            result = xpathexpression.evaluate(doc,XPathConstants.NODESET);
           Log.v(result.toString(), "Value of result");
    } catch (XPathExpressionException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
 NodeList nodelist =(NodeList) result;
 Node childlist = null;
  // org.w3c.dom.Element rootelement = doc.getDocumentElement();
  // NodeList nl = rootelement.getChildNodes();
   Log.v(new Integer(nodelist.getLength()).toString(), "Lenght of node list");

  for (int i =0; i< nodelist.getLength(); i++)
   {
      Node n = nodelist.item(i);
      System.out.println("message");
      childlist = n.getFirstChild();
      System.out.println("message2");
     String s = childlist.getNodeValue();  // THIS LINE GIVES EXCEPTION ELSE CODE IS              CORRECT

---

如果以下行导致NullPointerException，则childlist可能为null

String s = childlist.getNodeValue();

如果以下调用返回null，则对象childlist为null

childlist = n.getFirstChild();

如果n指向的对应xml节点中没有子节点，则类节点的方法getFirstChild（）将导致null。比较两种方法。因此，您应该检查xml节点n指向哪个节点，以及这个节点是否有child。为了得到更好的答案，我想您必须给出一个解析xml文档的示例

childlist = n.getFirstChild();