XMl 解析中的空指针异常

问题描述

我需要解析一个 Xml 文档并将值存储在文本文件中，当我解析普通数据时(如果所有标签都有数据)那么它工作正常，但如果任何标签没有数据那么它会抛出空指针异常"我需要做什么，以避免空指针异常，请用示例代码建议我示例 xml:

I need to parser an Xml Document and store the values in Text File, when i am parsing normal data (If all tags have data) then its working fine, but if any tag doesnt have data then it throwing "Null pointerException" what i need to do, to avoid null pointer exception, please suggest me with sample codes Sample xml:

<company>
    <staff>
        <firstname>John</firstname>
        <lastname>Kaith</lastname>
        <nickname>Jho</nickname>
        <Department>Sales Manager</Department>
    </staff>
    <staff>
        <firstname>Sharon</firstname>
        <lastname>Eunis</lastname>
        <nickname></nickname>
        <Department></Department>
    </staff>
    <staff>
        <firstname>Shiny</firstname>
        <lastname>mack</lastname>
        <nickname></nickname>
        <Department>SAP Consulting</Department>
    </staff>
</company>

代码:

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.w3c.dom.Document;
import org.w3c.dom.NodeList;
import org.w3c.dom.Node;
import org.w3c.dom.Element;
import java.io.File;

public class ReadXMLFile {

    public static void main(String argv[]) {

      try {

        File fXmlFile = new File("c:\file.xml");
        DocumentBuilderFactory dbFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
        Document doc = dBuilder.parse(fXmlFile);
        doc.getDocumentElement().normalize();

        System.out.println("Root element :" + doc.getDocumentElement().getNodeName());
        NodeList nList = doc.getElementsByTagName("staff");
        System.out.println("-----------------------");

        for (int temp = 0; temp < nList.getLength(); temp++) {

           Node nNode = nList.item(temp);
           if (nNode.getNodeType() == Node.ELEMENT_NODE) {

              Element eElement = (Element) nNode;

              System.out.println("First Name : " + getTagValue("firstname", eElement));
              System.out.println("Last Name : " + getTagValue("lastname", eElement));
                  System.out.println("Nick Name : " + getTagValue("nickname", eElement));
              System.out.println("Salary : " + getTagValue("Department", eElement));

           }
        }
      } catch (Exception e) {
        e.printStackTrace();
      }
  }

  private static String getTagValue(String sTag, Element eElement) {
    NodeList nlList = eElement.getElementsByTagName(sTag).item(0).getChildNodes();

        Node nValue = (Node) nlList.item(0);

    return nValue.getNodeValue();
  }

}

推荐答案

只要检查对象不是null:

private static String getTagValue(String tag, Element eElement) {
    NodeList nlList = eElement.getElementsByTagName(tag).item(0).getChildNodes();
    Node nValue = (Node) nlList.item(0);
    if(nValue == null) 
        return null;
    return nValue.getNodeValue();
}

String salary = getTagValue("Department", eElement);
if(salary != null) {
    System.out.println("Salary : " + getTagValue("Department", eElement));
}

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