Question

我使用SPARQL查询查询整数结果：

String qString = 
            "SELECT (COUNT(?S) AS ?C) "+
            "WHERE "+
            "{ "+
            "?S <"+p1+"> ?O ."+
            "?S <"+p2+"> ?O ."+
            "} ";
    Query q = QueryFactory.create(PREFIX+qString);
    QueryExecution qExecution = QueryExecutionFactory.sparqlService(ENDPOINT, q);
    ResultSet qResults = qExecution.execSelect();
    if(qResults.hasNext()){
        QuerySolution thisRow = qResults.next();
        int C = thisRow.get("C").toString();
    }

问题是get（）函数只返回使用toString（）的字符串输出。我需要一个int输出。我该怎么做呢？

Answer 1

我找到了解决方案。使用文字如下：

String qString = 
            "SELECT (COUNT(?S) AS ?C) "+
            "WHERE "+
            "{ "+
            "?S <"+p1+"> ?O ."+
            "?S <"+p2+"> ?O ."+
            "} ";
    Query q = QueryFactory.create(PREFIX+qString);
    QueryExecution qExecution = QueryExecutionFactory.sparqlService(ENDPOINT, q);
    ResultSet qResults = qExecution.execSelect();
    if(qResults.hasNext()){
        QuerySolution thisRow = qResults.next();
        Literal C_12_literal = ((Literal) thisRow.get("C"));
        C_12 = C_12_literal.getInt();
    }

Literal允许您获取许多数据类型，如int，float等。

Jena：如何从Java中的SPARQL（Jena）ResultSet获取Integer结果？

1 个答案: