SAX Parser for huge XML file

Hi.... if the XML file is small and repetitive then this will have meaning.. import javax.xml.parsers.; import org.xml.sax.; import org.xml.sax.helpers.; import java.io.;

public class SAXElementCount{  
  int startTag = 0;
  public static String ele;
  public static void main(String args[])throws IOException {
  BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
  System.out.print("Enter XML file name:");
  String xmlFile = bf.readLine();
  File file = new File(xmlFile);
  if (file.exists()){
  System.out.print("Enter XML tag name:");
  ele = bf.readLine();
  SAXElementCount tagCount = new SAXElementCount(xmlFile);
  }
  else{
  System.out.println("File not found!");
  }
  }
  public SAXElementCount(String str){
  try{
  SAXParserFactory parserFact = SAXParserFactory.newInstance();
  SAXParser parser = parserFact.newSAXParser();
  DefaultHandler dHandler = new DefaultHandler(){
  public void startElement(String uri, String name, String element, 
    Attributes atri)throws SAXException{
  if (element.equals(ele)){
  startTag++;
  }
  }
  public void endDocument(){
  System.out.println("Total elements: " + startTag);
  }
  };
  parser.parse(str,dHandler);
  }
  catch (Exception e){
  e.printStackTrace();
  }
  }
}

What if the xml file is very big or huge file? and sometimes we wont be aware of type of xml file...

View Answers

July 20, 2012 at 5:20 PM

Here is a code that parse xml file using dom parser in java.

import org.w3c.dom.*;
import org.w3c.dom.Node;
import javax.xml.parsers.*;
public class ParseXML{
    public static boolean isTextNode(Node n){
    return n.getNodeName().equals("#text");
  }
    public static void main(String[]args)throws Exception{
  DocumentBuilderFactory docFactory =  DocumentBuilderFactory.newInstance();
  DocumentBuilder docBuilder = docFactory.newDocumentBuilder();
  Document doc = docBuilder.parse("c:/file.xml");

          Element  element = doc.getDocumentElement(); 
          NodeList personNodes = element.getChildNodes();     
          for(int i=0; i<personNodes.getLength(); i++){
            Node emp = personNodes.item(i);
            if(isTextNode(emp))
            continue;
            NodeList nlist = emp.getChildNodes(); 

            for(int j=0; j<nlist.getLength(); j++ ){
            Node node = nlist.item(j);

            if(isTextNode(node)) 
              continue;

            System.out.print(node.getFirstChild().getNodeValue()+"\t ");
          } 
          System.out.println();
        }
    }
}

July 20, 2012 at 5:23 PM

I need to parse using sax parser... i wanna use direct file path... not by defining each and every element..

Thanks

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SAX Parser for huge XML file

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