以下のXMLファイルがあります
<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>
XML ファイルからデータを読み取る Java プログラムを以下に示します。
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");
int totalPersons = listOfPersons.getLength();
System.out.println("Total no of people : " + totalPersons);
for(int s=0; s<listOfPersons.getLength() ; s++){
Node firstPersonNode = listOfPersons.item(s);
if(firstPersonNode.getNodeType() == Node.ELEMENT_NODE){
Element firstPersonElement = (Element)firstPersonNode;
//-------
NodeList firstNameList = firstPersonElement.getElementsByTagName("first");
Element firstNameElement = (Element)firstNameList.item(0);
NodeList textFNList = firstNameElement.getChildNodes();
System.out.println("First Name : " +
((Node)textFNList.item(0)).getNodeValue().trim());
//-------
NodeList lastNameList = firstPersonElement.getElementsByTagName("last");
Element lastNameElement = (Element)lastNameList.item(0);
NodeList textLNList = lastNameElement.getChildNodes();
System.out.println("Last Name : " +
((Node)textLNList.item(0)).getNodeValue().trim());
//----
NodeList ageList = firstPersonElement.getElementsByTagName("age");
Element ageElement = (Element)ageList.item(0);
NodeList textAgeList = ageElement.getChildNodes();
System.out.println("Age : " +
((Node)textAgeList.item(0)).getNodeValue().trim());
//------
}//end of if clause
}//end of for loop with s var
}catch (SAXParseException err) {
System.out.println ("** Parsing error" + ", line "
+ err.getLineNumber () + ", uri " + err.getSystemId ());
System.out.println(" " + err.getMessage ());
}catch (SAXException e) {
Exception x = e.getException ();
((x == null) ? e : x).printStackTrace ();
}catch (Throwable t) {
t.printStackTrace ();
}
//System.exit (0);
}//end of main
}
そして結果は..
Root element of the doc is book
Total no of people : 3
First Name : Kiran
Last Name : Pai
Age : 22
First Name : Bill
Last Name : Gates
Age : 46
First Name : Steve
Last Name : Jobs
Age : 40
今私の質問は、このxmlを読むのに最速の他の方法があるかどうかアドバイスしてください、私はファストを探していました、アドバイスしてください..!!