Error in output of a simple SAX parser

帅比萌擦擦* 提交于 2019-12-13 03:44:28

问题


I saw an example in mykong - http://www.mkyong.com/java/how-to-read-xml-file-in-java-sax-parser/ I tried to make it work for the xml file (below) by making the following modifications to the code in the above page -

1 - Have only two if blocks in startElement() and characters() methods.
2 - Change the print statements in above methods, ie 
    FIRSTNAME and First Name = passenger id
    LASTNAME and Last Name = name        

The problem is - In the output, I see the word passenger instead of the value of passenger id. How do i fix that ?

<?xml version="1.0" encoding="utf-8"?>
<root xmlns:android="www.google.com">

<passenger id="001">
<name>Tom Cruise</name>
</passenger>

<passenger id="002">
<name>Tom Hanks</name>
</passenger>

</root>

Java Code

import javax.xml.parsers.SAXParser;
import javax.xml.parsers.SAXParserFactory;
import org.xml.sax.Attributes;
import org.xml.sax.SAXException;
import org.xml.sax.helpers.DefaultHandler;

public class ReadXMLFileSAX{

public static void main(String argv[]) {

try {

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

DefaultHandler handler = new DefaultHandler() {

boolean bfname = false;
boolean blname = false;

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

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

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

    if (qName.equalsIgnoreCase("name")) {
        blname = 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("passenger id : " + new String(ch, start,   length));
        bfname = false;
    }

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

}

 };

   saxParser.parse("c:\\flight.xml", handler);

 } catch (Exception e) {
   e.printStackTrace();
   }

   }

}

回答1:


In the startElement, when it's for "passenger", Attributes argument you get will have that value.

public void startElement(String uri, String localName,
    String qName, Attributes attributes)
        throws SAXException {
    if (qName.equalsIgnoreCase("passenger") && attributes != null){
        System.out.println(attributes.getValue("id"));
    }
}


来源:https://stackoverflow.com/questions/13798038/error-in-output-of-a-simple-sax-parser

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