Java SAX parsing: What's wrong with this XML?

Question

I'm trying to validate an XML file, but I get the following error:

Can not find declaration of element 'xsl:stylesheet'.

This is the XML:

<?xml version='1.0' encoding='utf-8'?>
<xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform' xmlns:msxsl='urn:schemas-microsoft-com:xslt' exclude-result-prefixes='msxsl' xmlns:ns='http://www.ibm.com/wsla'>
      <xsl:strip-space elements='*'/>
      <xsl:output method='xml' indent='yes'/>
      <xsl:template match='@* | node()'>
            <xsl:copy>
                  <xsl:apply-templates select='@* | node()'/>
            </xsl:copy>
      </xsl:template>
      <xsl:template match="/ns:SLA/ns:ServiceDefinition/ns:WSDLSOAPOperation/ns:SLAParameter/@name[.='TotalMemoryConsumption']">
            <xsl:attribute name='{name()}'>
                  <xsl:text>MemConsumption</xsl:text>
            </xsl:attribute>
      </xsl:template>
</xsl:stylesheet>

Where is the mistake?

EDIT: I want to parse this XML in Java with SAX, but I get the following error:

Element type "xsl:template" must be followed by either attribute specifications, ">" or "/>".

How to get rid of it?

Answer 1

Assuming you are actually trying to validate your XSL as an XML document, it looks like that website requires you to point to a schema or DTD in order to validate the XML against it. You can get a non-normative schema here: http://www.w3.org/TR/xslt20/#schema-for-xslt . Here's instructions on how to reference a schema from an XML file: http://www.ibm.com/developerworks/xml/library/x-tipsch.html

You could also check "Well-Formedness only," and check the document for well-formedness, if not actually validity.

Generally, any XSL engine will report any errors in your XSL document, so you don't need to validate it separately.

Answer 2

Your XSL is OK, don't worry. Just that there is no DTD/XSD for XSLs 1.0. no one bothers checking XSLT stylesheets (1.0) for validity. "Wellformedness" is enough.