SPARQL 构造。如何将 RDF 属性的值分配给实际的 RDF 变量

Question

我的查询（节选）大致是这样

CONSTRUCT {
?publication fb:type ?type;
fb:publicationType ?publicationType;
}
WHERE 
{
?publication a bibo:Document .
?publication vitro:mostSpecificType ?publicationType .
}

它returns输出类似于...

<rdf:Description rdf:about="https://abcd.fgh/individual/publication12345">
    <fb:publication>Example pub title</fb:publication>
    <fb:publicationType rdf:resource="http://purl.org/ontology/bibo/AcademicArticle"/>
</rdf:Description>

也许是初学者式的问题，但我如何调整查询以使输出为：

<rdf:Description rdf:about="https://abcd.fgh/individual/publication12345">
    <fb:publication>Example pub title</fb:publication>
    <fb:publicationType>Academic Article</fb:publicationType>
</rdf:Description>

谢谢

Answer 1

假设 bibo ontology is in the store you're querying, you can use its rdfs:label property in a property path:

CONSTRUCT {
  ?publication fb:type ?type;
    fb:publicationType ?publicationType;
} WHERE {
  ?publication a bibo:Document ;
    vitro:mostSpecificType/rdfs:label ?publicationType .
  FILTER (LANG(?publicationType) = "en")
}

对于?a vitro:mostSpecificType ?something. ?something rdfs:label ?b，?a vitro:mostSpecificType/rdfs:label ?b是shorthand，不绑定中间项。

SPARQL 构造。如何将 RDF 属性的值分配给实际的 RDF 变量

SPARQL Construct. How to assign the value of an RDF attribute to an actual RDF variable

rdf

ontology

sparql