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Re: [xsl] Namespace'd XPath expression in Java

2007-04-05 16:09:57
See below.
J.

Here is my Java code:
______________________________
// load XSD
DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
dbf.setNamespaceAware(true); <----------------------
DocumentBuilder docBuilder = dbf.newDocumentBuilder();
Document xsdDoc = docBuilder.parse("path to my xsd");

// eval XPath
XPath xpath = XPathFactory.newInstance().newXPath();
xpath.setNamespaceContext(new MyNamespaceContext());
System.out.println(xpath.evaluate("count(//xs:element)",
xsdDoc.getDocumentElement()));
______________________________

And finally implementation of NamespaceContext:

______________________________
public class MyNamespaceContext implements NamespaceContext {
        public String getNamespaceURI(String prefix) {
                if ("xs".equals(prefix)) {
                        return "http://www.w3.org/2001/XMLSchema";;
                } else if ("msdata".equals(prefix)) {
                        return "urn:schemas-microsoft-com:xml-msdata";
                }

                return XMLConstants.NULL_NS_URI;
        }

        public String getPrefix(String namespaceURI) {
                if ("http://www.w3.org/2001/XMLSchema".equals(namespaceURI)) {
                        return "xs";
                } else if 
("urn:schemas-microsoft-com:xml-msdata".equals(namespaceURI)) {
                        return "msdata";
                }

                return XMLConstants.DEFAULT_NS_PREFIX;
        }

        public Iterator<String> getPrefixes(String namespaceURI) {
                List<String> uris = new ArrayList<String>();
                uris.add(getPrefix(namespaceURI));
                return uris.iterator();
        }
}
______________________________

Thanks in advance.

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Jacek Radajewski

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