SPARQL查询“COUNT”在Virtuoso Jena API - QueryParseException

问题描述

相同的查询在DBpedia端点（）中有效，但不是在我的Java代码。
我只是试图使用COUNT函数提取频率。

Same query works in DBpedia Endpoint(http://ko.dbpedia.org/sparql), but not in my Java code.I am just trying to extract the frequency using "COUNT" function.

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)

我使用virt_jena2.jar和virtjdbc4.jar。
我已经通过类似的问题和答案（Jena ARQ扩展和SPARQL 1.1支持这个聚合查询 - 但我找不到如何更改它 - 我想我使用SPARQL1.1从事实，错误消息包括PARSERSPARQL11.java），但不能弄清楚如何解决这一点。

I am using virt_jena2.jar and virtjdbc4.jar.I have been looked through similar questions and answers(Jena ARQ extension and SPARQL 1.1 supports this aggregated query - But I couldn't find how to change it - I think I'm using SPARQL1.1 from the fact that error message includes PARSERSPARQL11.java ), but can't figure out how to solve this at this point.

提前感谢。

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();
}

推荐答案

这是非法的SPARQL语法：

This is illegal SPARQL syntax:

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

应该是

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

您将遇到？class in：

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

，因为它不是分组变量（使用 count 组的一切）。您是不是要有 GROUP BY？class ？

because it is not a grouped variable (using count you have a group of everything). Did you mean to have a GROUP BY ?class?

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