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我目前继耶拿API推理教程推理:基本RDFS与耶拿API
https://jena.apache.org/documentation/inference/
,并作为练习来测试我的理解,我想重写第一个例子,其从编程方式构建的模型证明了一个简单的RDFS推理:
import com.hp.hpl.jena.rdf.model.*;
import com.hp.hpl.jena.vocabulary.*;
public class Test1 {
static public void main(String...argv) {
String NS = "foo:";
Model m = ModelFactory.createDefaultModel();
Property p = m.createProperty(NS, "p");
Property q = m.createProperty(NS, "q");
m.add(p, RDFS.subPropertyOf, q);
m.createResource(NS + "x").addProperty(p, "bar");
InfModel im = ModelFactory.createRDFSModel(m);
Resource x = im.getResource(NS + "x");
// verify that property q of x is "bar" (which follows
// from x having property p, and p being a subproperty of q)
System.out.println("Statement: " + x.getProperty(q));
}
}
到一些东西,做相同,但与此龟文件,而不是读取模型(这是我自己的上述翻译,因此可能是越野车):
@prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>.
@prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>.
@prefix foo: <http://example.org/foo#>.
foo:p a rdf:Property.
foo:q a rdf:Property.
foo:p rdfs:subPropertyOf foo:q.
foo:x foo:p "bar".
与此代码:
public class Test2 {
static public void main(String...argv) {
String NS = "foo:";
Model m = ModelFactory.createDefaultModel();
m.read("foo.ttl");
InfModel im = ModelFactory.createRDFSModel(m);
Property q = im.getProperty(NS + "q");
Resource x = im.getResource(NS + "x");
System.out.println("Statement: " + x.getProperty(q));
}
}
这似乎不正确的方法(我怀疑,特别是我的q
财产的提取是有点不正确)。我究竟做错了什么?
这也正是它,非常感谢! – cjauvin 2013-04-05 01:29:18