1

我已经编写了代码来从我的 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();
    }
}

}

但现在我想提取与类关联的属性。我尝试使用 OntologyProperty 类但无法获得所需的输出

4

1 回答 1

0

你是什​​么意思extract the properties associated with the classes？

这可能意味着很多事情，你的意思是简单地找到具有该类作为域/范围的属性？

你不能使用 Ontology API 来做到这一点，它只是底层 API 的一个包装器，但你可以像这样使用较低级别的 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，您可以做同样的事情rdfs:range来获取将类作为范围的属性

于 2012-03-04T19:55:35.880 回答