使用ARQ(SPARQL)在Jena中进行java RDFS推理
这里的Jena文档https://jena.apache.org/documentation/javadoc/jena/index.html声明createOntologyModel
包含一个弱推理器,用于子类和子属性层次结构上的传递闭包,这就是我所寻找的。从一个简单的模型开始:
x:A rdf:type rdfs:Class .
x:A x:prop1 x:whatever .
x:B rdf:type rdfs:Class .
x:B rdfs:subClassOf x:A .
x:B x:prop2 x:other .
x:myInstance rdf:type x:B .
我试图查询rdf:type x:B
,并生成B
和超类A
的所有属性,例如
( ?all = <x:whatever> ) ( ?props = <x:prop1> ) -> ( ?type = <x:A> ) -> [Root]
( ?all = <x:A> ) ( ?props = <rdfs:subClassOf> ) -> ( ?type = <x:B> ) -> [Root]
( ?all = <x:other> ) ( ?props = <x:prop2> ) -> ( ?type = <x:B> ) -> [Root]
( ?all = <rdfs:Class> ) ( ?props = <rdf:type> ) -> ( ?type = <x:B> ) -> [Root]
我已经用这个例子做了一段时间的实验。它编译并运行,但只生成B的属性,不遍历subClassOf
树。我相信我错过了RDF模式的一个基本设置或使用,它使推理机能够完成自己的工作。有什么线索吗
public class jena2 {
private static void addRaw(OntModel m, String s, String p, String o) {
m.add(ResourceFactory.createStatement(
new ResourceImpl(s),new PropertyImpl(p),new ResourceImpl(o))
);
}
public static void main(String[] args) {
OntModel model = ModelFactory.createOntologyModel();
addRaw(model, "x:A", "rdf:type", "rdfs:Class");
addRaw(model, "x:A", "x:prop1", "x:whatever");
addRaw(model, "x:B", "rdf:type", "rdfs:Class");
addRaw(model, "x:B", "x:prop2", "x:other");
addRaw(model, "x:B", "rdfs:subClassOf", "x:A");
addRaw(model, "x:widget", "rdf:type", "x:B");
StringBuffer sb = new StringBuffer();
sb.append("PREFIX x: <x:>");
sb.append("PREFIX rdf: <rdf:>");
sb.append("PREFIX rdfs: <rdfs:>");
sb.append("SELECT *");
sb.append("WHERE {");
sb.append(" x:widget rdf:type ?type .");
sb.append(" ?type ?props ?all .");
sb.append("}");
Query query = QueryFactory.create(sb.toString());
try (QueryExecution qexec = QueryExecutionFactory.create(query, model)) {
ResultSet results = qexec.execSelect() ;
for ( ; results.hasNext() ; ) {
QuerySolution soln = results.nextSolution() ;
System.out.println(soln);
}
} catch(Exception e) {
System.out.println("epic fail: " + e);
}
}
# 1 楼答案
(这不是完整的答案!)
根据上面的评论,一些工作代码:
输出:
它还返回RDFS类
A
的属性