[Solved] To write a query SPARQL in Java code using strstarts filter

Question

I have resolved my problem using this code:

public QueryExecution query(){

        String stringa = "http://dbpedia.org/resource/Fred_Guy";

        ParameterizedSparqlString qs = new ParameterizedSparqlString( "" +
                "prefix dbpediaont: <http://dbpedia.org/ontology/>\n" +
                "prefix dbpedia: <http://dbpedia.org/resource/>\n" +
                "prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>\n" +
                "\n" +  
                "select ?resource where {\n" +
                "?mat rdf:type ?resource\n" +
                "filter strstarts(str(?resource), dbpediaont:)\n" +
                "}" );


        Resource risorsa = ResourceFactory.createResource(stringa);
        qs.setParam( "mat", risorsa );

        //System.out.println( qs );

        QueryExecution exec = QueryExecutionFactory.sparqlService( "http://dbpedia.org/sparql", qs.asQuery() );


        ResultSet results = ResultSetFactory.copyResults( exec.execSelect() );

        while ( results.hasNext() ) {

            System.out.println( results.next().get( "resource" ));
        }

        // A simpler way of printing the results.
        ResultSetFormatter.out( results );

        return exec;

    }

In particular, in FILTER as the second arguments, I have written the “dbpediaont” label.

This code, however, still produces an error:

Exception in thread "main" HttpException: 500
    at com.hp.hpl.jena.sparql.engine.http.HttpQuery.execGet(HttpQuery.java:340)
    at com.hp.hpl.jena.sparql.engine.http.HttpQuery.exec(HttpQuery.java:276)
    at com.hp.hpl.jena.sparql.engine.http.QueryEngineHTTP.execSelect(QueryEngineHTTP.java:345)

This error is due to the fact that the code need to write

dbpediaont:

with

str()

since it’s an IRI, not a string:

filter strstarts(str(?resource), str(dbpediaont:))

The latter problem is reported on HttpException error when I call SPARQL query (on DBPedia) in Java Code

I hope this answer can help someone.

Accepted Answer