Using Saxon --> Understanding parsing and transforming

ratnalein · June 17th, 2013, 11:57 AM

Hallo everyone,

my company uses Saxon to process their xml-files. I am trying to understand what kind of parsing method they are using:

Code:

System.setProperty("javax.xml.transform.TransformerFactory","net.sf.saxon.TransformerFactoryImpl");
...
CharArrayWriter buffer = new CharArrayWriter();
StreamResult resultSource = new StreamResult(buffer);
trans.transform(xmlSource, resultSource);
csv = buffer.toString();

I got an explanation, which I could not really understand. My question was: where can I see that we have parsed the source xml-file before we transform it.
The explanation was:
"The parsing is done "on the fly". So, a DOM-Tree or others was not created before."

Where from the above code is meant?

Other questions:
1. What is the default parser from Saxon?
2. Is my assumtion correct, by calling the method getTransformer, the xml-file is being parsed implizitly?

Thank you very much

Regards,

Ratna

mhkay · June 17th, 2013, 12:03 PM

You haven't said what is in xmlSource, and this is rather critical to your question. It's likely that it's one of the three "standard" kinds of Source defined by JAXP:

StreamSource - in this case Saxon parses the supplied XML stream using the default SAX parser, and builds a tree in Saxon's native format (a TinyTree) ready for use in the transformation.

SAXSource - in this case, if the XmlReader of the SAXSource is initialized, it works the same way as a StreamSource except that the XmlReader supplies the parser that Saxon uses.

DOMSource - in this case a pre-built tree in DOM format is supplied, so Saxon does not need to do any parsing. This is less efficient because Saxon works much faster with its native tree format than with an external DOM.

ratnalein · June 17th, 2013, 06:09 PM

Hallo Mr. Kay,

yes, indeed the xmlSource is from type StreamSource. So, the Default SAX Parser is used in this case. Just to be sure:
so, there is no "parsing"-method needed for the parsing process? The parsing process is implicitely implemented by calling the method Transformer.transform ?

Thank you.

Regards,

Ratna

mhkay · June 17th, 2013, 06:53 PM

>The parsing process is implicitely implemented by calling the method Transformer.transform ?

Yes.

By the way, let me recommend my book. I think it will help your understanding.