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Exceptions in SAX parser code

how the arguments are used that i am passing through the "SAXException" constructor in the following example.

catch (IOException e) {

throw new SAXException("I/O error", e);

}

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    author's profile photo Former Member
    Former Member
    Posted on Nov 30, 2007 at 08:45 AM

    Hi Sankar,

    how the arguments can be used while passing through SAXException, this you can learn very well in Java tutorials provided by sun java.

    below are the link helpful to you these will describe you how to use arguments and also other help that might be useful for you.

    1)This package provides the <b>core SAX APIs</b>.

    http://java.sun.com/javase/6/docs/api/org/xml/sax/package-summary.html

    2)This package contains interfaces to <b>SAX2 facilities</b> that conformant SAX drivers won't necessarily support.

    http://java.sun.com/javase/6/docs/api/org/xml/sax/ext/package-summary.html

    3)This package contains <b>"helper" classes</b>, including support for <b>bootstrapping SAX-based applications</b>.

    http://java.sun.com/javase/6/docs/api/org/xml/sax/helpers/package-summary.html

    hope this may solve your problem

    Thanks

    sandeep Sharma

    PS: if helpful please reward points

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  • Posted on Nov 30, 2007 at 07:28 AM
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  • author's profile photo Former Member
    Former Member
    Posted on Nov 30, 2007 at 08:53 AM

    Hello Sankar,

    Please frame the Question more properly.I didnt underestand what you have written.Could you pls re construct your Query

    With Regards

    Vidya

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