[英]RDFS: How to subclass a FOAF class
Is it possible to subclass a class found in FOAF ( http://xmlns.com/foaf/spec/ )? 是否可以继承在FOAF( http://xmlns.com/foaf/spec/ )中找到的类? I tried something like the code below, but I am not sure if it is the proper way to do it or not.
我尝试了类似以下代码的方法,但是我不确定这是否是正确的方法。
<rdfs:Class rdf:ID="user">
<rdfs:subClassOf rdf:resource="http://xmlns.com/foaf/0.1/#Agent" />
<rdfs:comment>
The class of users, subclass of foaf:Agent.
</rdfs:comment>
</rdfs:Class>
Your snippet, although not a complete RDF document, is the right way to make yourdoc#user
a subclass of http://xmlns.com/foaf/0.1/#Agent
. 您的代码段虽然不是完整的RDF文档,但却是使
yourdoc#user
成为http://xmlns.com/foaf/0.1/#Agent
子类的正确方法。 However, that latter class is not the FOAF Agent class. 但是,后一类不是FOAF代理类。 The FOAF Agent class is identified by the URI
http://xmlns.com/foaf/0.1/Agent
(with no #
). FOAF代理类由URI
http://xmlns.com/foaf/0.1/Agent
(不带#
)标识。 It might be useful to have a look at the actual FOAF ontology , because you can see how it defines subclasses of Agent. 查看实际的FOAF本体可能很有用,因为您可以看到它如何定义Agent的子类。 For instance, it declared foaf:Organization with
例如,它声明了foaf:Organization with
<rdfs:Class rdf:about="http://xmlns.com/foaf/0.1/Organization" rdfs:label="Organization" rdfs:comment="An organization." vs:term_status="stable">
<rdf:type rdf:resource="http://www.w3.org/2002/07/owl#Class"/>
<rdfs:subClassOf rdf:resource="http://xmlns.com/foaf/0.1/Agent"/>
<rdfs:isDefinedBy rdf:resource="http://xmlns.com/foaf/0.1/"/>
<owl:disjointWith rdf:resource="http://xmlns.com/foaf/0.1/Person"/>
<owl:disjointWith rdf:resource="http://xmlns.com/foaf/0.1/Document"/>
</rdfs:Class>
If you're writing this by hand, it's much easier to work in the Turtle or N3 serialization, where that would be: 如果您是手工编写的,那么在Turtle或N3序列化中工作起来会容易得多,这将是:
foaf:Organization a rdfs:Class , owl:Class ;
rdfs:comment "An organization." ;
rdfs:isDefinedBy foaf: ;
rdfs:label "Organization" ;
rdfs:subClassOf foaf:Agent ;
owl:disjointWith foaf:Person , foaf:Document ;
vs:term_status "stable" .
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