Sax Parser: processingInstruction() not called

Question

Hi,
maybe this is a n00b-question.

I try to parse an xml-file like that:

<?xml version="1.0" encoding="UTF-8" ?>
<test>
  <a></a>
</test>

with the following code:

public static void parse(File f) {
    final DefaultHandler handler = new DefaultHandler() {
        @Override
        public void processingInstruction(String target, String data) throws SAXException {
            System.out.println("Processing Instruction");
        }
    };
    SAXParserFactory.newInstance().newSAXParser().parse(f, handler);
}

I expect the output "Processing Instruction" to be printed on stdout. But this doesn't happen. Can somebody tell me why?

Answer 1

From JavaDoc:

A SAX parser must never report an XML declaration (XML 1.0, section 2.8) or a text declaration (XML 1.0, section 4.3.1) using this method.

As I understand it, the <?xml version="1.0" encoding="UTF-8" ?> line will not be reported/catched by this method. Try to add another processing instruction in the xml code and feed it to the parser again.