java XML 解析中的空指针异常
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Null Pointer Exception in XMl Parsing
提问by Jean Logeart
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:
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，但是如果任何标签没有数据那么它会抛出我需要的“空指针异常”要做，为了避免空指针异常，请建议我使用示例代码示例 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>
code:
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();
}
}
回答by Jean Logeart
Just check that the object is not null:
只需检查对象不是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));
}
回答by JB Nizet
NodeList nlList = eElement.getElementsByTagName(sTag).item(0).getChildNodes();
The above line gets the child nodes of the element with the given tag name. If this element doesn't have data, the returned node list is empty.
上面的行获取具有给定标签名称的元素的子节点。如果该元素没有数据，则返回的节点列表为空。
Node nValue = (Node) nlList.item(0);
The above line gets the first element from the empty node list. Since the list is empty, 0 is an invalid index, and as per the documentation, null is returned
上面的行从空节点列表中获取第一个元素。由于列表为空，0 是无效索引，根据文档，返回 null
return nValue.getNodeValue();
The above line thus calls a method on a null variable, which causes the NPE.
因此，上面的行调用了一个空变量上的方法，这会导致 NPE。
You should test if the list is empty, and return what you want (an empty string, for example), if it's the case:
您应该测试列表是否为空，并返回您想要的内容（例如，空字符串），如果是这样：
if (nList.getLength() == 0) {
return "";
}
Node nValue = (Node) nlList.item(0);
return nValue.getNodeValue();
回答by Massimiliano Giunchi
This code should work:
此代码应该工作：
// getNode function
private static String getNode(String sTag, Element eElement) {
//if sTag exists(not null) I get childNodes->nlList
if (eElement.getElementsByTagName(sTag).item(0)!=null){
NodeList nlList = eElement.getElementsByTagName(sTag).item(0).getChildNodes();
//check if child (nlList) contain something
if ((nlList.getLength() == 0))//if the content is null
return "";
//if child contains something extract the value of the first element
Node nValue = (Node) nlList.item(0);
return nValue.getNodeValue();
}
return "";
}
回答by Sungray
Replace
代替
System.out.println("First Name : " + getTagValue("firstname", eElement));
by
经过
System.out.println("First Name : " + getTagValue("firstname", eElement) == null?"":getTagValue("firstname", eElement));
same thing for the other tags.
其他标签也一样。
