我有以下SPARQL查询:
SELECT * FROM {
?measurement a oboe-core:Measurement ;
oboe-core:ofCharacteristic oboe-core:Name ;
oboe-core:usesStandard :Anaee-franceExperimentalSiteNamingStandard ;
oboe-core:hasValue ?anaeeSiteNameStandard .
BIND ( IRI( CONCAT( "http://www.anaee-france.fr/ontology/anaee-france_ontology" ,
?anaeeSiteNameStandard ) ) AS ?site ) .
OPTIONAL { ?site rdfs:label ?_anaeeSiteName . }
BIND ( IF (BOUND (?_anaeeSiteName), ?_anaeeSiteName, "NULL_anaeeSiteName"@en ) AS
?anaeeSiteName) .
FILTER (lang( ?anaeeSiteName ) = "en") .
} limit 3
_anaeeSiteName为空,知道我的图形包含:
<http://www.anaee-france.fr/ontology/anaee-france_ontology#Guyaflux>
rdfs:label "Guyaflux"@en .
当我直接使用时
BIND ( IRI( "http://www.anaee-france.fr/ontology/anaee-france_ontology#Guyaflux" ) AS ?site ) .
代替BIND IRI CONCAT
BIND ( IRI( CONCAT( "http://www.anaee-france.fr/ontology/anaee-france_ontology" ,
?anaeeSiteNameStandard ) ) AS ?site ) .
我得到了一些结果。
有人能告诉我怎么了吗?
谢谢
我们同意Uniformated和Andy的观点,认为存在语法错误。您缺少FROM和WHERE关键字的图形uri,并且需要concat中的#来匹配您的示例uri。
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对不起,我删除了"FROM";以便简化查询。
完整的Sparql查询如下:
SELECT * FROM <foret> WHERE {
?measurement a oboe-core:Measurement ;
oboe-core:ofCharacteristic oboe-core:Name ;
oboe-core:usesStandard :Anaee-franceExperimentalSiteNamingStandard ;
oboe-core:hasValue ?anaeeSiteNameStandard .
BIND ( IRI( CONCAT( "http://www.anaee-france.fr/ontology/anaee-france_ontology" ,
?anaeeSiteNameStandard ) ) AS ?site ) .
OPTIONAL { ?site rdfs:label ?_anaeeSiteName . }
BIND ( IF (BOUND (?_anaeeSiteName), ?_anaeeSiteName, "NULL_anaeeSiteName"@en ) AS
?anaeeSiteName) .
FILTER (lang( ?anaeeSiteName ) = "en") .
} limit 3
奇怪的是,当我添加:时
BIND ( str(?anaeeSiteNameStandard) AS ?anaeeSiteNameStandard ) .
之前:
BIND ( IRI( CONCAT( "http://www.anaee-france.fr/ontology/anaee-france_ontology" ,
?anaeeSiteNameStandard ) ) AS ?site ) .
我得到了预期的结果!