How can I get parent node while i using SAX parser?

Question

I need to parse document using SAX parser in java. I was able to print all the node values if I use DefaultHandler class traditionally implementing the startElement, endElement and characters method. How can we access the the previous node value at child node, how can I do that?
My Sample XML is:

<staff>
        <firstname>yong</firstname>
        <lastname>mook kim</lastname>
        <nickname>mkyong</nickname>
        <salary>100000</salary>
    </staff>
    <staff>
        <firstname>low</firstname>
        <lastname>yin fong</lastname>
        <nickname>fong fong</nickname>
        <salary>200000</salary>
    </staff>

Based on salary node value, I also want to access the first name. I am confused. How can we do it? My sample Code:

SAXParserFactory factory = SAXParserFactory.newInstance();
    SAXParser saxParser = factory.newSAXParser();

    DefaultHandler handler = new DefaultHandler() {

    boolean bfname = false;
    boolean blname = false;
    boolean bnname = false;
    boolean bsalary = false;

    public void startElement(String uri, String localName,String qName, 
                Attributes attributes) throws SAXException {

        System.out.println("Start Element :" + qName);

        if (qName.equalsIgnoreCase("FIRSTNAME")) {
            bfname = true;
        }

        if (qName.equalsIgnoreCase("LASTNAME")) {
            blname = true;
        }

        if (qName.equalsIgnoreCase("NICKNAME")) {
            bnname = true;
        }

        if (qName.equalsIgnoreCase("SALARY")) {
            bsalary = true;
        }

    }

    public void endElement(String uri, String localName,
        String qName) throws SAXException {

        System.out.println("End Element :" + qName);

    }

    public void characters(char ch[], int start, int length) throws SAXException {

        if (bfname) {
            System.out.println("First Name : " + new String(ch, start, length));
            bfname = false;
        }

        if (blname) {
            System.out.println("Last Name : " + new String(ch, start, length));
            blname = false;
        }

        if (bnname) {
            System.out.println("Nick Name : " + new String(ch, start, length));
            bnname = false;
        }

        if (bsalary) {
            //System.out.println("Salary : " + new String(ch, start, length));
                    String nodeValue=new String(ch, start, length);
                    if(nodeValue.compareTo("100000")==0)
                   {
                     **????I need to store the respective respective first name
                         in ArrayList**
                    }
            bsalary = false;
        }

    }

     };

Aravind Yarram · Accepted Answer · 2011-12-27 13:12:21Z

3

You can't navigate back and forth when using SAX. You should try using DOM. If you have to use SAX then you can use Stack to hold the previous data and pop them as required.

answered Dec 27, 2011 at 13:12

Aravind Yarram

80.5k49 gold badges239 silver badges335 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

mevada.yogesh · Accepted Answer · 2011-12-27 13:15:57Z

1

You can use a String variable to store the name as

public void characters(char ch[], int start, int length) throws SAXException {
    ... Code Here ...       
    if (bfname) {
        employeeName = new String(ch, start, length);
        bfname = false;
    }
    ... Code Here ...
}

& use this variable at the end as

    public void characters(char ch[], int start, int length) throws SAXException {
    ... Code Here ...
    if (bsalary) {
        String nodeValue=new String(ch, start, length);\
            if(nodeValue.compareTo("100000")==0)
            {
                //Use employeeName Here...
            }
        bsalary = false;
    }
    ... Code Here ...
}

answered Dec 27, 2011 at 13:15

mevada.yogesh

1,1383 gold badges12 silver badges37 bronze badges

Collectives™ on Stack Overflow

How can I get parent node while i using SAX parser?

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related