SPARQL构造.如何将RDF属性的值分配给实际的RDF变量 [英] SPARQL Construct. How to assign the value of an RDF attribute to an actual RDF variable

问题描述

我的查询(摘录)大致是这个.

My query (excerpt) is roughly this.

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

它返回类似于...的输出

And it returns output similar to...

<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>

也许是一个初学者式的问题，但是我如何调整查询以使输出为:

Perhaps a beginner-esque question, but how do I tweak the query such that the output is:

<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

推荐答案

假设 bibo本体在您要查询的商店中，您可以在

?a vitro:mostSpecificType/rdfs:label ?b是?a vitro:mostSpecificType ?something. ?something rdfs:label ?b的简写，没有绑定中间术语.

The ?a vitro:mostSpecificType/rdfs:label ?b is shorthand for ?a vitro:mostSpecificType ?something. ?something rdfs:label ?b, without binding the intermediate term.

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