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读取xml的最快和最优化的方法

我有以下XML文件

<book>
<person>
  <first>Kiran</first>
  <last>Pai</last>
  <age>22</age>
</person>
<person>
  <first>Bill</first>
  <last>Gates</last>
  <age>46</age>
</person>
<person>
  <first>Steve</first>
  <last>Jobs</last>
  <age>40</age>
</person>
</book>
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现在,从XML文件读取数据的Java程序如下所示.

import java.io.File;
import org.w3c.dom.Document;
import org.w3c.dom.*;

import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.DocumentBuilder;
import org.xml.sax.SAXException;
import org.xml.sax.SAXParseException; 

public class ReadAndPrintXMLFile{

    public static void main (String argv []){
    try {

            DocumentBuilderFactory docBuilderFactory = DocumentBuilderFactory.newInstance();
            DocumentBuilder docBuilder = docBuilderFactory.newDocumentBuilder();
            Document doc = docBuilder.parse (new File("book.xml"));

            // normalize text representation
            doc.getDocumentElement ().normalize ();
            System.out.println ("Root element of the doc is " + 
                 doc.getDocumentElement().getNodeName());


            NodeList listOfPersons = doc.getElementsByTagName("person"); …
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java xml

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2
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