> The result of a SPARQL query contains only what you selected in the > SPARQL query. Nothing more, nothing less.
So how should I change the query to get a property value? Thanks Dmitri Pisarenko 08.02.2017, 10:21, "Lorenz B." <[email protected]>: > The result of a SPARQL query contains only what you selected in the > SPARQL query. Nothing more, nothing less. And you select only "things > that have an email address". > >> Hello! >> >> I have the following code, in which I find a resource by its e-mail. >> >> val varn = "x" >> val query = createQuery("""SELECT ?${varn} >> WHERE { ?x >> <http://www.w3.org/2001/vcard-rdf/3.0#EMAIL> "${email}" }""") >> val qexec = createQueryExecution(ds, query) >> val rs = qexec.execSelect() >> if (rs.hasNext()) { >> val solution = rs.nextSolution() >> val rec = solution[varn] >> // Here I need to find the value of the property >> FirstContactTime >> } >> >> Now I want to find out whether `rec` has a property `FirstContactTime` and >> if yes, its value. >> >> I tried >> `rec.model.listObjectsOfProperty(ds.defaultModel.createProperty(FirstContactTime))` >> but it doesn't return anything. The debugger says `rec` does have a >> property `FirstContactTime`. >> >> How can I get the value of `FirstContactTime` (`2017-03-03T10:35:00Z`) in >> my code? >> >> Thanks in advance >> >> Dmitri Pisarenko > -- > Lorenz Bühmann > AKSW group, University of Leipzig > Group: http://aksw.org - semantic web research center
