Java XPath: Queries with default namespace xmlns

Question

I want to do an XPath query on this file (excerpt shown):

<?xml version="1.0" encoding="UTF-8"?>
<!-- MetaDataAPI generated on: Friday, May 25, 2007 3:26:31 PM CEST -->
<ModelClass xmlns="http://xml.sap.com/2002/10/metamodel/webdynpro" xmlns:IDX="urn:sap.com:WebDynpro.ModelClass:2.0">
    <ModelClass.Parent>
        <Core.Reference package="com.test.mypackage" name="ModelName" type="Model"/>

This is a snippet of the code I'm using:

DocumentBuilderFactory domFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = domFactory.newDocumentBuilder();
Document document = builder.parse(new File(testFile));
XPathFactory factory = XPathFactory.newInstance();
XPath xpath = factory.newXPath();
xpath.setNamespaceContext( new NamespaceContext() {
    public String getNamespaceURI(String prefix) {
...

String result = xpath.evaluate(xpathQueryString, document);
System.out.println(result);

The problem I'm facing is that when the default namespace is referenced in an XPath query, the getNamespaceURI method is not called to resolve it. This query for example doesn't extract anything:

//xmlns:ModelClass.Parent/xmlns:Core.Reference[@type=\"Model\"]/@package

Now I've tried "tricking" the parser by replacing xmlns with a fake prefix d and then writing the getNamespaceURI method accordingly (so to return http://xml.sap.com/2002/10/metamodel/webdynpro when d is encountered). In this case, the getNamespaceURI is called but the result of the XPath expression evaluation is always an empty string.

If I strip out namespaces from the file and from the XPath query expression, I can get the string I wanted (com.test.mypackage).

Is there a way to make things work properly with the default namespace?

score 13 · Answer 1 · answered May 23 '12 at 16:19

The XPath 1.0 specification requires that "no prefix means no namespace". So JAXP, which was designed for XPath 1.0, is quite right to stop you binding the "null prefix" to some non-null namespace.

XPath 2.0 allows you to declare a default namespace for unqualified names in your XPath expression, but to take advantage of that you will need an API (such as Saxon's s9api) that takes advantage of this feature.

Martin Honnen · Accepted Answer · 2012-05-23T13:32:29.540

11

In your Namespace context, bind a prefix of your choice (e.g. df) to the namespace URI in the document

xpath.setNamespaceContext( new NamespaceContext() {
    public String getNamespaceURI(String prefix) {
      switch (prefix) {
        case "df": return "http://xml.sap.com/2002/10/metamodel/webdynpro";
        ...
       }
    });

and then use that prefix in your path expressions to qualify element names e.g. /df:ModelClass/df:ModelClass.Parent/df:Core.Reference[@type = 'Model']/@package.

edited May 23 '12 at 13:32

answered May 23 '12 at 13:26

Martin Honnen

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6

Make sure you also create a namespace aware DocumentBuilder http://docs.oracle.com/javase/6/docs/api/javax/xml/parsers/DocumentBuilderFactory.html#setNamespaceAware%28boolean%29 before working with namespaces. – Martin Honnen May 23 '12 at 14:08
That did the trick. I thought that the DocumentBuilder was by default aware of namespaces. Thank you again. – Pietro M. May 23 '12 at 14:13

Java XPath: Queries with default namespace xmlns

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