Re: Matching literals of unknown langs

[Andy Seaborne]

On 17/12/12 11:41, Olivier Rossel wrote:
Hello.
The SPARQL spec says:
"Florence" is not the same RDF literal as "Florence"@fr

and it's just repeating RDF.

...

Now I have to federate-query an italian dataset and the french dbPedia.

As seen above, my french dbPedia contains this literal: "Florence"@fr
My italian dataset contains this literal: "Florence" (with no lang tag).

Here is the federated query:
SELECT DISTINCT ?LocalityITA ?LocalityFR WHERE {
  SERVICE <http://91.121.14.47:6665/sparql/> {
     ?Address <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>
<http://www.w3.org/2006/vcard/ns#Address> .
     ?Address <http://www.w3.org/2006/vcard/ns#locality> ?LocalityITA .
     ?LocalityITA <http://www.w3.org/2000/01/rdf-schema#label> ?LabelITA .
  }
  BIND (strbefore(?LabelITA, "(") AS ?Label)
  SERVICE <http://fr.dbpedia.org/sparql>{
     ?LocalityFR <http://www.w3.org/2000/01/rdf-schema#label> ?Label
  }
}

Reformatted:

SELECT DISTINCT  ?LocalityITA ?LocalityFR
WHERE
  { SERVICE <http://91.121.14.47:6665/sparql/>
      { ?Address rdf:type vc:Address .
        ?Address vc:locality ?LocalityITA .
        ?LocalityITA rdfs:label ?LabelITA
      }
    BIND(strbefore(?LabelITA, "(") AS ?Label)
    SERVICE <http://fr.dbpedia.org/sparql>
      { ?LocalityFR rdfs:label ?Label }
  }


You can use str() to get just the lexical form:


How can I tune the query so the literal matching works across lang tags?
Thanks for your help.

You could canonicalise to the simple literal in each SERVICE

SERVICE <....> {
   ....
   ?Address vc:locality ?l .
    BIND(str(?l) AS ?locality)
}



Except fr.dbpedia.org/Virtuoso does not support BIND.

You can get the same effect with a subquery:

SERVICE <....> {
   SELECT (str(?l as ?locality)
   { ...
      ?Address vc:locality ?l .
   }
}

Now you have ?locality without a language tag and can use it as the 
canonical term (if yoru app thinks that's safe enough).

        Andy