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   RE: [xml-dev] SAXParser error

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Title: Message
The String argument to parse is not the XML content, it is the URL of the resource containing the content.
Try creating a StringReader and create an InputSource that wraps the Reader.

Michael Kay
Software AG
home: Michael.H.Kay@ntlworld.com
work: Michael.Kay@softwareag.com

-----Original Message-----
From: Barry Lulas [mailto:Barry.Lulas@BristolWest.com]
Sent: 08 November 2002 15:00
To: 'xml-dev@lists.xml.org'
Subject: [xml-dev] SAXParser error

I have a web service that receives a string containing well-formed XML.  I am trying to parse this string using the SAXParser and the default handler. 

        // This is the client side:
        StringBuffer buff = new StringBuffer();
        buff.append("<?xml version='1.0' encoding='UTF-8'?>");
        buff.append("<myelement attr1 = 'false' attr2 = 'true'>");
        buff.append("  <nestedelement>value</nestedelement>");
        String s = new String(buff);

        // This is the server side that receives the string reqeust
        DefaultHandler handler = new DefaultHandler();
                SAXParserFactory factory = SAXParserFactory.newInstance();
                SAXParser parser = factory.newSAXParser();

                // The exception comes when I call parse, trying to parse the string as an XML document
                parser.parse(s, handler);
        catch (Exception e)
                // The exception I'm getting is:  no protocol: PRINTS OUT THE XML TEXT

Anyone know what the source of the exception is?

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