CODEWITHSUNDEEP
rdfsparqlontology
abent
abent

Reputation: 319

Sparql query with uri

I have rdf structure like this

<owl:Thing rdf:about="http://hust.se.vtio.owl#atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi">
<rdf:type rdf:resource="http://hust.se.vtio.owl#ATM"/>
<rdfs:label xml:lang="vn"><![CDATA[ATM - Techcombank]]></rdfs:label>
<rdfs:label xml:lang="en"><![CDATA[ATM - Techcombank]]></rdfs:label>
<hasLatitude rdf:datatype="&xsd;double">20.9954529</hasLatitude>
<hasLongtitude rdf:datatype="&xsd;double">105.8546176</hasLongtitude>
<hasGeoPoint rdf:datatype="http://franz.com/ns/allegrograph/3.0/geospatial/spherical/degrees/-180.0/180.0/-90.0/90.0/5.0">+20.9954529+105.8546176</hasGeoPoint>
<hasLocation rdf:resource="http://hust.se.vtio.owl#atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi-address"/>
<belongToBank rdf:resource="http://hust.se.vtio.owl#techcombank"/>
<hasMedia rdf:resource="http://hust.se.vtio.owl#atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi-images"/>

How can I get label and Latitude... by sparql when I know uri :

http://hust.se.vtio.owl#atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi

Upvotes: 0

Views: 317

Answers (1)

Artemis
Artemis

Reputation: 3301

It depends on how you have defined the ontology. For example, let's imagine that you have defined something like x subClassOf: hasLatitude value 20.9954529, then you can ask a similar query to the one below:

prefix :<http://hust.se.vtio.owl#>
SELECT  *
    WHERE { ?s rdfs:label ?label.
            ?s rdfs:subClassOf  ?o.
            ?o owl:onProperty :hasLatitude.
            ?o ?x ?y.
 }

You can filter ?s to only give you answers for atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi. For example, filter (?s=:atm-techcombank-127-pho-minh-khai-hai-ba-trung-ha-noi).

Upvotes: 1

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