Your problem is the default namespace. Check out this article for how to deal with namespaces in your XPath: http://www.edankert.com/defaultnamespaces.html
One of the conclusions they draw is:
So, to be able to use XPath
expressions on XML content defined in
a (default) namespace, we need to
specify a namespace prefix mapping
Note that this doesn't mean that you have to change your source document in any way (though you're free to put the namespace prefixes in there if you so desire). Sounds strange, right? What you will do is create a namespace prefix mapping in your java code and use said prefix in your XPath expression. Here, we'll create a mapping from spreadsheet to your default namespace.
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
// there's no default implementation for NamespaceContext...seems kind of silly, no?
xpath.setNamespaceContext(new NamespaceContext() {
public String getNamespaceURI(String prefix) {
if (prefix == null) throw new NullPointerException("Null prefix");
else if ("spreadsheet".equals(prefix)) return "http://schemas.openxmlformats.org/spreadsheetml/2006/main";
else if ("xml".equals(prefix)) return XMLConstants.XML_NS_URI;
return XMLConstants.NULL_NS_URI;
}
// This method isn't necessary for XPath processing.
public String getPrefix(String uri) {
throw new UnsupportedOperationException();
}
// This method isn't necessary for XPath processing either.
public Iterator getPrefixes(String uri) {
throw new UnsupportedOperationException();
}
});
// note that all the elements in the expression are prefixed with our namespace mapping!
XPathExpression expr = xpath.compile("/spreadsheet:workbook/spreadsheet:sheets/spreadsheet:sheet[1]");
// assumin
answered 2011-06-17T18:58:25.437