Question

为了试验SWRL语言，我运行了（在Neatbeans 8.2中）以下简单的java代码：

String base = "http://www.prova/testont.owl";
IRI ontologyIRI = IRI.create(base);
OWLOntologyManager manager = OWLManager.createOWLOntologyManager();
OWLOntology ontology = manager.createOntology(ontologyIRI);
OWLDataFactory factory = manager.getOWLDataFactory();

OWLClass adult = factory.getOWLClass(IRI.create(ontologyIRI + "#Adult"));
OWLClass person = factory.getOWLClass(IRI.create(ontologyIRI + "#Person"));
OWLDataProperty hasAge = factory.getOWLDataProperty(IRI.create(ontologyIRI + "#hasAge"));

OWLNamedIndividual john = factory.getOWLNamedIndividual(IRI.create(ontologyIRI + "#John"));
OWLNamedIndividual andrea = factory.getOWLNamedIndividual(IRI.create(ontologyIRI + "#Andrea"));

OWLClassAssertionAxiom classAssertion = factory.getOWLClassAssertionAxiom(person, john);
manager.addAxiom(ontology, classAssertion);
classAssertion = factory.getOWLClassAssertionAxiom(person, andrea);
manager.addAxiom(ontology, classAssertion);

OWLDatatype integerDatatype = factory.getOWLDatatype(OWL2Datatype.XSD_INTEGER.getIRI());
OWLLiteral literal = factory.getOWLLiteral("41", integerDatatype);
OWLAxiom ax = factory.getOWLDataPropertyAssertionAxiom(hasAge, andrea, literal);
manager.addAxiom(ontology, ax);

literal = factory.getOWLLiteral("15", integerDatatype);
ax = factory.getOWLDataPropertyAssertionAxiom(hasAge, john, literal);
manager.addAxiom(ontology, ax);

SWRLRuleEngine ruleEngine = SWRLAPIFactory.createSWRLRuleEngine(ontology);
ruleEngine.createSWRLRule("r1", "Person(?p)^hasAge(?p,?age)^swrlb:greaterThan(?age,17) -> Adult(?p)");

manager.saveOntology(ontology, IRI.create(((new File("FILE_PATH")).toURI())));

我使用了具有以下依赖关系的maven：

 <dependency>
     <groupId>edu.stanford.swrl</groupId>
     <artifactId>swrlapi-drools-engine</artifactId>
     <version>1.1.4</version>
  </dependency>

我收到以下错误：

线程“main”中的异常org.swrlapi.parser.SWRLParseException：无效的SWRL原子谓词'Person' at org.swrlapi.parser.SWRLParser.generateEndOfRuleException（SWRLParser.java:479）在org.swrlapi.parser.SWRLParser.parseSWRLAtom（SWRLParser.java:210） at org.swrlapi.parser.SWRLParser.parseSWRLRule（SWRLParser.java:106） at org.swrlapi.factory.DefaultSWRLAPIOWLOntology.createSWRLRule（DefaultSWRLAPIOWLOntology.java:219） at org.swrlapi.factory.DefaultSWRLAPIOWLOntology.createSWRLRule（DefaultSWRLAPIOWLOntology.java:213） at org.swrlapi.factory.DefaultSWRLRuleAndQueryEngine.createSWRLRule（DefaultSWRLRuleAndQueryEngine.java:249） at ilc.cnr.it.swrl4morphology.SimpleToSWRL.main（SimpleToSWRL.java:450）

但是，如果我将ontology保存在文件中然后重新加载它，我就不会再收到错误了。似乎在首次保存期间添加了默认前缀。这听起来很奇怪......

拜托，你能帮我理解我的错误吗？

提前致谢，安德烈

Answer 1

在保存和解析时，使用本体IRI作为基础，将相对IRI（如“Person”）转换为绝对IRI。

SWRL规则在OWLAPI本体中创建

1 个答案: