Question

我在Jena模型上有一个返回单个结果的SPARQL查询。如何才能访问该结果，因为只有一个元素，我无法迭代？我尝试了2个选项，但都失败了。我使用ResultSetFormatter将结果转换为JSONObject，但我发现键不是我的变量。此外，我尝试使用toList（）方法将其转换为QuerySolution列表，但它返回空列表。有什么帮助吗？

public void insertMedcationContext(JSONObject medcontext) {

    connection.getDataset().begin(ReadWrite.WRITE);
    Model model = connection.getDataset().getDefaultModel();

    String medActivityQuery = "PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>\n"
            + "PREFIX fn: <http://www.w3.org/2005/xpath-functions#>\n"
            + "PREFIX medication:<http://www.cs.kaist.ac.kr/medication/ontology#>\n"
            + "PREFIX resource:<http://www.cs.kaist.ac.kr/medication/resource#>\n"
            + "PREFIX time:<http://www.w3.org/2006/time#>\n"
            + "SELECT ?activity ((?deschour - ?timestamp) AS ?gap) (fn:abs(?gap)AS ?gapabsolute)\n"
            + "WHERE\n"
            + "{?activity rdf:type medication:MedicationActivity .\n"               
            + "?activity medication:belongsTo ?schedule .\n"
            + "?activity medication:expectedTime ?time .\n"
            + "?time time:hasTimeDescription ?desc .\n"
            + "?desc time:year ?descyear .\n"
            + "?desc time:month ?descmonth .\n"
            + "?desc time:day ?descdate .\n"
            + "?desc time:hour ?deschour .\n"
            + "}\n"
            + "ORDER BY (?gapabsolute)\n"
            + "LIMIT 1";              

    try {
        Resource schedule = model.createResource(
                nameSpace + medcontext.getString("schedule"),
                MEDICATION.Schedule);
        Resource scheduleResource = model.getResource(schedule.getURI());

        ParameterizedSparqlString parameterizedQuery = new ParameterizedSparqlString(
                medActivityQuery);
        parameterizedQuery.setParam("schedule", scheduleResource);

        JSONObject timestamp = new JSONObject();
        timestamp = medcontext.getJSONObject("exacttime");
        parameterizedQuery.setLiteral("descyear",Integer.toString(timestamp.getInt("year")),XSDDatatype.XSDgYear);
        parameterizedQuery.setLiteral("descmonth",Integer.toString(timestamp.getInt("month")),XSDDatatype.XSDgMonth);
        parameterizedQuery.setLiteral("descdate",Integer.toString(timestamp.getInt("date")),XSDDatatype.XSDgDay);
        parameterizedQuery.setLiteral("timestamp",Integer.toString(timestamp.getInt("hour")),XSDDatatype.XSDnonNegativeInteger);

        Query query = QueryFactory.create(parameterizedQuery.toString());
        QueryExecution qe = QueryExecutionFactory.create(query, model);

        try {
            ResultSet result = qe.execSelect();
            String text = ResultSetFormatter.asText(result);
            System.out.println(text);

            ByteArrayOutputStream b = new ByteArrayOutputStream();
            ResultSetFormatter.outputAsJSON(b, result);
            JSONObject jsonResult = new JSONObject (b.toString());  

            System.out.print(jsonResult);

            List <QuerySolution> resultList = ResultSetFormatter.toList(result);                

            // Get the right medication activity from the model for which context is incoming   
                Resource rightActivity = null;  

                QuerySolution row = resultList.get(0); 
                rightActivity = row.getResource("activity");
                System.out.print(rightActivity.toString());

resultList为空，但有一个结果......

Answer 1

默认情况下，ResultSet是按需生成的。您只能迭代它们一次，然后消耗结果。你做完

之后

ResultSet result = qe.execSelect();
String text = ResultSetFormatter.asText(result);

你可能无法从

获得任何结果

ResultSetFormatter.outputAsJSON(b, result);

或

ResultSetFormatter.toList(result);

相反，您应该使用例如

复制ResultSet

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

如何从Jena SPAQL Resultset获取第一个元素

1 个答案: