2013-04-04 58 views
0

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?

回答

1
String NS = "foo:"; 
m.createResource(NS + "x") 

创建一个URI,但龟版有富:X = http://example.org/foo#x

通过打印模式im.write(System.out, "TTL");

变化NS = "foo:"见分歧NS = "http://example.org/foo#"

+0

这也正是它,非常感谢! – cjauvin 2013-04-05 01:29:18