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Namespace prefix not declared error after extracting a node in OWL/XML file with Java & xPath

Tags:

java

xml

xpath

owl

Initially I have this file.

<rdf:RDF xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#">
    <owl:Class />
    <owl:Class />
    <owl:ObjectProperty />
    <Situation:Situation rdf:about"http://localhost/rdf#situa0">
        <Situation:composedBy />
    </Situation:Situation>
</rdf:RDF>

My goal is to extract the node Situation and its content using xPath "RDF/Situation" ...

<Situation:Situation rdf:about"http://localhost/rdf#situa0">
    <Situation:composedBy />
</Situation:Situation>

I found a good example to work with in Java How to extract a complete XML block.

I changed names of tags to my own since I use namespaces and predefined tags.

Here's my code

 public static void main(String... args) throws Exception {
        String xml = "<rdf:RDF xmlns:rdf=\"http://www.w3.org/1999/02/22-rdf-syntax-ns#\"><owl:Class /><owl:Class /><owl:ObjectProperty /><Situation:Situation rdf:about=\"http://localhost/rdf#situa0\" ><Situation:composedBy /></Situation:Situation></rdf:RDF>";
        DocumentBuilderFactory dbf = DocumentBuilderFactory.newInstance();
        Document doc = dbf.newDocumentBuilder().parse(
                new InputSource(new StringReader(xml)));

        XPath xPath = XPathFactory.newInstance().newXPath();
        Node result = (Node) xPath.evaluate("RDF/Situation", doc, XPathConstants.NODE);

        System.out.println(nodeToString(result));
    }

    private static String nodeToString(Node node) throws TransformerException {
        StringWriter buf = new StringWriter();
        Transformer xform = TransformerFactory.newInstance().newTransformer();
        xform.setOutputProperty(OutputKeys.OMIT_XML_DECLARATION, "yes");
        xform.transform(new DOMSource(node), new StreamResult(buf));
        return (buf.toString());
    }

My goal is 90% achieved but I have a problem, the Situation tag has an attribute about with a prefix rdf (the code works if I remove the prefix, and even if I added rdf xmlns in the root element)

<Situation:Situation rdf:about="http://localhost/rdf#situa0">

I got this error

ERROR: 'The namespace prefix' rdf 'has not been declared.' Exception in thread "main" javax.xml.transform.TransformerException: java.lang.RuntimeException: Namespace prefix 'rdf' has not been declared. com.sun.org.apache.xalan.internal.xsltc.trax.TransformerImpl.transform at (Unknown Source) com.sun.org.apache.xalan.internal.xsltc.trax.TransformerImpl.transform at (Unknown Source)

I added dbf.setNamespaceAware(true) as @ Ian Roberts mentioned, so I got other errors asking for owl & Situation namespaces, after adding it in the root tag, I got nothing in output and without errors. What is the problem ?? The problem was that the variable result, this time is null, so there's a problem with the xPath query..

I tried to see in another place the result of the query and it worked fine in an online xPath tester.

enter image description here

So what is the problem ??

Is there any other way to do like this job.???

thx :)

like image 866
Ali Ben Messaoud Avatar asked Apr 16 '26 14:04

Ali Ben Messaoud


1 Answers

Is there any other way to do like this job?

Yes, there are other,more appropriate, ways to do this job.

It's typically not a great idea to try to process RDF documents using XML tools, since the same RDF graph can often be represented a number of different ways in RDF/XML. This is discussed in more detail in my answer to How to access OWL documents using XPath in Java?, but we can see the issue pretty quickly here. After adding some additional namespace declarations your data looks like this:

<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:Situation="https://stackoverflow.com/q/22170071/1281433/"
    xmlns:owl="http://www.w3.org/2002/07/owl#">
  <owl:Class/>
  <owl:Class/>
  <owl:ObjectProperty/>
  <Situation:Situation rdf:about="http://localhost/rdf#situa0">
    <Situation:composedBy></Situation:composedBy>
  </Situation:Situation>
</rdf:RDF>

The same RDF graph can be serialized like this, too:

<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:Situation="https://stackoverflow.com/q/22170071/1281433/"
    xmlns:owl="http://www.w3.org/2002/07/owl#" > 
  <rdf:Description rdf:nodeID="A0">
    <rdf:type rdf:resource="http://www.w3.org/2002/07/owl#Class"/>
  </rdf:Description>
  <rdf:Description rdf:about="http://localhost/rdf#situa0">
    <rdf:type rdf:resource="https://stackoverflow.com/q/22170071/1281433/Situation"/>
    <Situation:composedBy></Situation:composedBy>
  </rdf:Description>
  <rdf:Description rdf:nodeID="A1">
    <rdf:type rdf:resource="http://www.w3.org/2002/07/owl#ObjectProperty"/>
  </rdf:Description>
  <rdf:Description rdf:nodeID="A2">
    <rdf:type rdf:resource="http://www.w3.org/2002/07/owl#Class"/>
  </rdf:Description>
</rdf:RDF>

If you're looking for a Situation:Situation element, you'll find one in the first serialization, but not the second, even though they're the same RDF graph.

You could probably use a SPARQL query to get what you're looking for. The typical implementation of describe queries might do what you want. E.g., the very simple query

describe <http://localhost/rdf#situa0>

produces this result (in RDF/XML):

<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:Situation="https://stackoverflow.com/q/22170071/1281433/"
    xmlns:owl="http://www.w3.org/2002/07/owl#">
  <Situation:Situation rdf:about="http://localhost/rdf#situa0">
    <Situation:composedBy></Situation:composedBy>
  </Situation:Situation>
</rdf:RDF>

Alternatively, you could ask for everything that has the type Situation:Situation:

prefix s: <https://stackoverflow.com/q/22170071/1281433/>
describe ?situation where {
  ?situation a s:Situation .
}
<rdf:RDF
    xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
    xmlns:s="https://stackoverflow.com/q/22170071/1281433/"
    xmlns:owl="http://www.w3.org/2002/07/owl#">
  <s:Situation rdf:about="http://localhost/rdf#situa0">
    <s:composedBy></s:composedBy>
  </s:Situation>
</rdf:RDF>

The important point here is to use an appropriate query language for the type of data that you have. You have RDF, which is a graph-based data representation. An RDF graph is a set of triples. Your data is five triples:

_:BX2D6970b66dX3A1448f4e1bcfX3AX2D7ffe <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <http://www.w3.org/2002/07/owl#Class> .
<http://localhost/rdf#situa0> <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <https://stackoverflow.com/q/22170071/1281433/Situation> .
<http://localhost/rdf#situa0> <https://stackoverflow.com/q/22170071/1281433/composedBy> "" .
_:BX2D6970b66dX3A1448f4e1bcfX3AX2D7ffd <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <http://www.w3.org/2002/07/owl#ObjectProperty> .
_:BX2D6970b66dX3A1448f4e1bcfX3AX2D7fff <http://www.w3.org/1999/02/22-rdf-syntax-ns#type> <http://www.w3.org/2002/07/owl#Class> .

In the Turtle serialization, the graph is:

@prefix owl:   <http://www.w3.org/2002/07/owl#> .
@prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix Situation: <https://stackoverflow.com/q/22170071/1281433/> .

[ a       owl:Class ] .

<http://localhost/rdf#situa0>
        a                     Situation:Situation ;
        Situation:composedBy  "" .

[ a       owl:Class ] .

[ a       owl:ObjectProperty ] .

You should use SPARQL (the standard RDF query language) or an RDF-based API for extracting data from RDF documents.

like image 122
Joshua Taylor Avatar answered Apr 18 '26 04:04

Joshua Taylor



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