[英]SPARQL Query “COUNT” in Virtuoso Jena API - QueryParseException
Same query works in DBpedia Endpoint( http://ko.dbpedia.org/sparql ), but not in my Java code. 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。 I am just trying to extract the frequency using "COUNT" function.
我只是想使用“ COUNT”功能提取频率。
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();
And the error is as follows. 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)
I am using virt_jena2.jar and virtjdbc4.jar. 我正在使用virt_jena2.jar和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.
我一直在寻找类似的问题和答案(Jena ARQ扩展和SPARQL 1.1支持此聚合查询-但我找不到如何更改它-我认为由于错误消息包含PARSERSPARQL11,我正在使用SPARQL1.1。 java),但目前还无法解决。
Thanks in advance. 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();
}
This Jena code (without Virtuoso) gives me same error message. 这个耶拿代码(没有Virtuoso)给了我同样的错误信息。
This is illegal SPARQL syntax: 这是非法的SPARQL语法:
SELECT ... count(distinct ?s) as ?count where
It should be 它应该是
SELECT ... (count(distinct ?s) as ?count) where
The you will have a problem with ?class
in: 您会在
?class
中遇到问题:
SELECT ?class (count(distinct ?s) as ?count) where
because it is not a grouped variable (using count
you have a group of everything). 因为它不是一个分组变量(使用
count
您拥有一组所有内容)。 Did you mean to have a GROUP BY ?class
? 您是说要有
GROUP BY ?class
?
关键字为“ Encountered " "count" "COUNT ""
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