如何在SPARQL中将日期值(xsd:date)与年份进行比较？

Question

我创建了一个SPARQL查询来比较年份dbo:birthDate是否与以下值不同dbo:birthYear:

PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX dbo: <http://dbpedia.org/ontology/>
prefix xsd: <http://www.w3.org/2001/XMLSchema#>
prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#>  

Select distinct ?s ?vv1 ?vv2 where
  { ?s a dbo:Person ; dbo:birthDate ?v1 .
    ?s dbo:birthYear ?v2 . 
    bind(year(xsd:date(?v1)) as ?vv1) .
    bind(xsd:gYear(?v2) as ?vv2) .
    filter(?vv1 !=  ?vv2)
  }

我得到了以下答案:

                S                  vv1    vv2
<http://example.com/resource/1>   1993   1993
<http://example.com/resource/5>   1998   1993   
<http://example.com/resource/7>   1978   1978   
<http://example.com/resource/13>  1993   1993 

正确的答案只是资源5,即年份birthDate等于1998和年份birthYear等于1993.答案显示,vv1并且vv2没有多年的价值" ",显然两者都有相同的类型.因此,我不明白为什么SPARQL表示这一点,vv1并且例如,vv2记录中1是不同的.我查找了如何比较SPARQL中的日期,我认为我的查询是正确的.

我使用Fuseki 2.3.1测试了这个查询

我的实例示例:

@prefix dbo:   <http://dbpedia.org/ontology/> .
@prefix :      <http://example.com/resource/> .
@prefix dbp:   <http://dbpedia.org/property/> .
@prefix owl:   <http://www.w3.org/2002/07/owl#> .
@prefix rdf:   <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
@prefix xml:   <http://www.w3.org/XML/1998/namespace> .
@prefix xsd:   <http://www.w3.org/2001/XMLSchema#> .
@prefix mu:    <http://example.com/resource/> .
@prefix rdfs:  <http://www.w3.org/2000/01/rdf-schema#> .
@prefix foaf:  <http://xmlns.com/foaf/0.1/> .

mu:1    a                dbo:Actor , 
                         dbo:Person , 
                         dbo:MusicalArtist , 
                         mu:MusicActor , 
                         owl:NamedIndividual ;
        dbo:birthDate    "1993-03-11"^^xsd:date ;
        dbo:birthYear    "1993" ;
        foaf:name        "John Joe"@en , 
                         "Joe, John"@en .

mu:5    a                <http://example.com/MusicActor> , 
                         dbo:Actor , 
                         dbo:Person , 
                         mu:MusicActor , 
                         owl:NamedIndividual ;
        dbo:birthDate    "1998-04-11" , 
                         "1998-04-11"^^xsd:date ;
        dbo:birthYear    "1993" ;
        foaf:name        "Barry Hannah"@en , 
                         "Barry B. Hannah"@en .

mu:7    a                dbo:Person , 
                         dbo:MusicalArtist , 
                         owl:NamedIndividual ;
        dbo:birthDate    "1978-05-11" , 
                         "1978-05-11"@en ;
        dbo:birthYear    "1978" ;
        foaf:name        "Helen Petty" , 
                         "Helen Petty"@en .

请问,有人可以说这是为什么会这样吗？遗憾的是我不知道.任何帮助将非常感谢!

Answer 1

SPARQL year()函数返回一个整数,因此您需要强制?v2转换为整数:

SPARQL DISTINCT ?s ?vv1 ?vv2
WHERE {
   ?s a dbo:Person ; dbo:birthDate ?v1 .
   ?s dbo:birthYear ?v2 . 
   BIND(year(xsd:date(?v1)) as ?vv1) .
   BIND(xsd:integer(?v2) as ?vv2) .
   FILTER(?vv1 != ?vv2)
}