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
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Michael.H.Kay@n...
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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>");
buff.append("</myelement>");
String
s = new String(buff);
// This
is the server side that receives the string reqeust
DefaultHandler handler = new DefaultHandler();
...
try
{
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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