Question

我已编写代码从我的RDF文件中提取类和子类..这是代码..我正在使用dotNetRDf库..

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System;
using System.Linq;
using VDS.RDF;
using VDS.RDF.Ontology;
using VDS.RDF.Parsing;


namespace ConsoleApplication1
{
class Program
{
    static void Main(string[] args)
    {
    //
        OntologyGraph g = new OntologyGraph();
        FileLoader.Load(g, "D:\\SBIRS.owl");
        OntologyClass someClass = g.CreateOntologyClass(new    
        Uri("http://www.semanticweb.org/ontologies/2012/0/SBIRS.owl#Shape"));

                  //Write out Super Classes

        foreach (OntologyClass c in someClass.SuperClasses)
        {
           Console.WriteLine("Super Class: " + c.Resource.ToString());
        }
        //Write out Sub Classes



        foreach (OntologyClass c in someClass.SubClasses)
        {

            Console.WriteLine("Sub Class: " + c.Resource.ToString());
        }
        Console.Read();
    }
}

}

但是现在我想提取与类相关的属性..我试图使用OntologyProperty类但是无法获得所需的输出

Answer 1

extract the properties associated with the classes是什么意思？

这可能意味着任何数量的东西，你的意思是简单地找到将该类作为域/范围的属性吗？

你不能使用Ontology API来做这个，它只是底层API的包装，但是你可以像这样使用低级API：

//Assuming you've already set up your Graph and Class as above...

//Find properties who have this class as a domain
INode domain = g.CreateUriNode(new Uri(NamespaceMapper.RDFS + "domain"));
IEnumerable<OntologyProperty> ps = g.GetTriplesWithPredicateObject(domain, someClass).Select(t => new OntologyProperty(t.Subject, g));

//Now iterate over ps and do what you want with the properties

同样，您可以使用rdfs:range执行相同的操作，以获取将类作为范围的属性

从RDF文件中提取与类关联的属性

1 个答案: