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