SPARQL select optional with language

Devin McQueeney picture Devin McQueeney · Oct 29, 2011 · Viewed 11.2k times · Source

I have some triples that look like this:

test:thing rdfs:label "Non-Language Label"
test:thing rdfs:label "English Label"@en
test:thing rdfs:label "French Label"@fr

I'd like to form a sparql query that gives me the "Non-Language Label" AND the "French Label", if any exists.

I tried this and it's not working:

PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> 
SELECT ?label ?preferredLabel
WHERE {
   test:thing rdfs:label ?label 
   OPTIONAL {
     test:thing rdfs:label ?preferredLabel . 
     FILTER (regex(str(?preferredLabel), '(^|\\\\W)fr', 'i'))
   }
}

Thanks in advance!

Answer

RobV picture RobV · Oct 29, 2011

I don't see why you need OPTIONAL here at all. Jan's query is failing because there is no shared variable between the outer pattern and the optional so you are trying to calculate the cross product of every label for test:thing with every non/french labelled test:thing which may be huge and why the query processor is failing.

You simply want something like the following unless I've misunderstood your question

PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> 
SELECT ?label
WHERE 
{
   test:thing rdfs:label ?label 
   FILTER(LANG(?label) = "" || LANGMATCHES(LANG(?label), "fr"))
}

If you want the two labels separately then you could do something like:

PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> 
SELECT ?label ?preferredLabel
WHERE 
{
  {
   test:thing rdfs:label ?label . FILTER(LANG(?label) = "")
  }
  UNION
  {
   test:thing rdfs:label ?preferredLabel . FILTER(LANGMATCHES(LANG(?label), "fr"))
  }
}