parser.parse()

[steve francisco]

The API shows SAXParser.parse() method has to take 2 arguments, one is the File or InputSource and the 2nd one is the DefaultHandler. But in many samples I saw "parser.parse(file)" or "parser.parse(inputSource)" without the 2nd argument in the method. why ?

[Bajji Pat]

It is because, mostly we use ParserAdapter and not the SAXParser as such, and we use PA's parse method that takes an InputSource or a systemId. We do it because ParserAdapter wraps SAX1 and makes it act as a SAX2 ,with feature, property, and Namespace support etc..

Hope this helps

Balaji
[ May 20, 2005: Message edited by: Bajji Pat ]