XMl解析中的空指针异常

  • Z时代
  • 2024-01-10
  • 分类：问答

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（如果所有标签都包含数据）时，它的工作状况很好，但是如果任何标签中都没有数据，则它会抛出“

NullpointerException”这样做，为避免出现异常" title="空指针异常">空指针异常，请使用示例代码Sample xml来建议我：

<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>

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();
  }
}

回答：

只需检查对象是否不是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));
}

以上是 XMl解析中的空指针异常 的全部内容， 来源链接： utcz.com/qa/414171.html

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