Question

相同的查询在DBpedia Endpoint（http://ko.dbpedia.org/sparql）中有效，但在我的Java代码中却无效。我只是想用＆＃34; COUNT＆＃34;来提取频率。功能

VirtGraph set = new VirtGraph("http://ko.dbpedia.org", HOST, USERNAME, PASSWORD);
Query freqsparql = QueryFactory.create("SELECT ?class count(distinct ?s) as ?count where{?s <http://ko.dbpedia.org/property/이름> ?o. ?s a ?class.} order by DESC(?count)");
VirtuosoQueryExecution freqvqe = VirtuosoQueryExecutionFactory.create(freqsparql, set);
ResultSet freqresults = freqvqe.execSelect();

错误如下。

Exception in thread "main" com.hp.hpl.jena.query.QueryParseException: Encountered " "count" "count "" at line 1, column 15.
Was expecting one of:
<VAR1> ...
<VAR2> ...
"from" ...
"where" ...
"(" ...
"{" ...

at com.hp.hpl.jena.sparql.lang.ParserSPARQL11.perform(ParserSPARQL11.java:102)
at com.hp.hpl.jena.sparql.lang.ParserSPARQL11.parse$(ParserSPARQL11.java:53)
at com.hp.hpl.jena.sparql.lang.SPARQLParser.parse(SPARQLParser.java:37)
at com.hp.hpl.jena.query.QueryFactory.parse(QueryFactory.java:148)
at com.hp.hpl.jena.query.QueryFactory.create(QueryFactory.java:80)
at com.hp.hpl.jena.query.QueryFactory.create(QueryFactory.java:53)
at com.hp.hpl.jena.query.QueryFactory.create(QueryFactory.java:41)

我正在使用virt_jena2.jar和virtjdbc4.jar。我已经查看了类似的问题和答案（Jena ARQ扩展和SPARQL 1.1支持这个聚合查询 - 但我无法找到如何更改它 - 我想我从错误的事实中使用SPARQL1.1消息包括PARSERSPARQL11.java），但此时无法弄清楚如何解决这个问题。

提前致谢。

String sparqlQueryString = "SELECT ?class count(distinct ?s) as ?count    where{?s <http://ko.dbpedia.org/property/이름> ?o. ?s a ?class.} order by DESC(?count)";
Query query = QueryFactory.create(sparqlQueryString);
QueryExecution qexec = QueryExecutionFactory.sparqlService(
                "http://ko.dbpedia.org/sparql", query);
try {
    ResultSet results = qexec.execSelect();
    while(results.hasNext()){
        QuerySolution freqresult = results.nextSolution();
        RDFNode domain = freqresult.get("class");
        RDFNode freqcount = freqresult.get("count");
        System.out.println(freqresult);
        System.out.println(domain + "---" + freqcount);
    }
} catch (Exception e) {
    e.printStackTrace();
} finally {
    qexec.close();
}

这个Jena代码（没有Virtuoso）给了我同样的错误信息。

Answer 1

这是非法的SPARQL语法：

      1 23 4

应该是

SELECT ... count(distinct ?s) as ?count where

您将遇到SELECT ... (count(distinct ?s) as ?count) where的问题：

?class

因为它不是分组变量（使用SELECT ?class (count(distinct ?s) as ?count) where你有一组所有东西）。你的意思是count吗？

Answer 2

关键字为g.setFont(new Font("TimesRoman", Font.PLAIN, fontSize)); g.setColor(Color.red); g.drawString("some string", 40, 40);

SPARQL查询＆＃34; COUNT＆＃34;在Virtuoso Jena API中 - QueryParseException

2 个答案: