问题是： 我希望这段代码具有特殊Individual的rdf：type属性，但是model.write方法给了我这样的输出：
<rdf:RDF
xmlns:rdf="http://www.w3.org/1999/02/22-rdf-syntax-ns#"
xmlns:j.0="http://microhard.com/graph/"
xmlns:owl="http://www.w3.org/2002/07/owl#"
xmlns:j.1="http://microhard.com/property/"
xmlns:rdfs="http://www.w3.org/2000/01/rdf-schema#"
xmlns:xsd="http://www.w3.org/2001/XMLSchema#">
<owl:Class rdf:about="http://microhard.com/graph/POI"/>
<j.0:POI>
<j.1:p1>111</j.1:p1>
</j.0:POI>
</rdf:RDF>
这是代码：
public class Main {
private static final String NS = "http://microhard.com/graph/";
public static void main(String[] args) {
OntModel m = ModelFactory.createOntologyModel(OntModelSpec.OWL_MEM);
OntClass klass = m.createClass(NS + "POI");
Individual in = klass.createIndividual();
in.addProperty(m.createProperty("http://microhard.com/property/p1"), "111");
m.write(System.out, "RDF/XML");
}
}