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Nov 16th, 2018
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SPARQL 0.68 KB | None | 0 0
  1. SELECT ?p ?label (COUNT(?s) AS ?COUNT) WHERE { {?s ?p ?o FILTER REGEX(STR(?p), "http://www.profium.com/city/") . ?p rdfs:label ?label FILTER LANGMATCHES( LANG(?label), "fi" ) . ?s <http://www.profium.com/archive/depictedObjectInverse> ?objectInverse} UNION { ?s ?p ?o FILTER REGEX(STR(?p),  "http://www.profium.com/tuomi") . ?p rdfs:label ?label FILTER LANGMATCHES( LANG(?label), "fi" ) . ?s <http://www.profium.com/archive/depictedObjectInverse> ?objectInverse} UNION { ?s ?p ?o FILTER REGEX(STR(?p), "http://www.profium.com/tuomitos") . ?p rdfs:label ?label FILTER LANGMATCHES( LANG(?label), "fi" ) . ?s <http://www.profium.com/archive/depictedObjectInverse> ?objectInverse} }  GROUP BY ?p ?label
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