Question

我想获取作为参数传递的Job ID，并在owl文件中检查带有id的id，然后使该Job ID的工作技能与应聘者的技能相匹配（候选人也包含技能）。我正在获取url参数并将其存储在变量中。

String jobID = request.getParameter("JobID");

String query =
                "PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>" +
                        "PREFIX owl: <http://www.w3.org/2002/07/owl#> " +
                        "PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> " +
                        "PREFIX ns0: <http://www.Jobsurveillance.com/CV#> " +
                        "SELECT (str(?object) as ?label) " +
                        "(str(?object2) as ?label2) " +
                        "(str (?phoneNumobj) as ?label3) " +
                        "(str (?degreeobj) as ?label4) " +
                        "(str (?jobcatObj) as ?label5) " +
                        "(str (?candidateSkill) as ?label8) " +
                        "(str (?jobAdID) as ?label6) " +
                        "(str (?jobSkill) as ?label7) WHERE { " +
                        " ?person ns0:Full_Name ?object ." +
                        " ?person ns0:Email_Id ?object2 ." +
                        " ?person ns0:Phone_Number ?phoneNumobj ." +
                        " ?person ns0:Degree ?degreeobj ." +
                        " ?person ns0:Job_Category ?jobcatObj ." +
                        " ?person ns0:Skills ?candidateSkill ." +
                        " ?job ns0:JobID ?jobAdID ." +
                        " ?job ns0:JobSkills ?jobSkill ." +
                        " filter (regex(?jobAdID, \"" + jobID + "\") ." +
                        " filter (?jobSkill = ?candidateSkill)) ." +
                        "}";

我运行时收到如下错误：

com.hp.hpl.jena.query.QueryParseException: Encountered " "." ". "" at line 1, 
column 756.
Was expecting one of:
"not" ...
"in" ...
<INTEGER_POSITIVE> ...
<DECIMAL_POSITIVE> ...
<DOUBLE_POSITIVE> ...
<INTEGER_NEGATIVE> ...
<DECIMAL_NEGATIVE> ...
<DOUBLE_NEGATIVE> ...
")" ...
"=" ...
"!=" ...
">" ...
"<" ...
"<=" ...
">=" ...
"||" ...
"&&" ...
"+" ...
"-" ...
"*" ...
"/" ...

我的数据集如下

<owl:NamedIndividual rdf:about="http://www.Jobsurveillance.com/CV#Candidate6">
  <rdf:type>
    <owl:Restriction>
      <owl:onProperty rdf:resource="http://www.Jobsurveillance.com/CV#includes"/>
      <owl:someValuesFrom rdf:resource="http://www.Jobsurveillance.com/CV#CV"/>
    </owl:Restriction>
  </rdf:type>

  <ns0:Address rdf:datatype="http://www.w3.org/2001/XMLSchema#string">Colombo 15</ns0:Address>
  <ns0:Company rdf:datatype="http://www.w3.org/2001/XMLSchema#string">TechOrin</ns0:Company>
  <ns0:Degree rdf:datatype="http://www.w3.org/2001/XMLSchema#string">B.Sc (Hons), Information Technology</ns0:Degree>
  <ns0:Email_Id rdf:datatype="http://www.w3.org/2001/XMLSchema#string">yashmitha@gmail.com</ns0:Email_Id>
  <ns0:Skills rdf:datatype="http://www.w3.org/2001/XMLSchema#string">Java</ns0:Skills>
  <ns0:Job_Category rdf:datatype="http://www.w3.org/2001/XMLSchema#string">Marketing</ns0:Job_Category>
  <ns0:Full_Name rdf:datatype="http://www.w3.org/2001/XMLSchema#string">Yashmitha</ns0:Full_Name>
  <ns0:Phone_Number rdf:datatype="http://www.w3.org/2001/XMLSchema#integer">0723679017</ns0:Phone_Number>
</owl:NamedIndividual>

这是我的SPARQL查询。我是否以正确的方式实现了过滤器。任何人都可以帮我这个忙。谢谢！

Answer 1

问题已解决。下面的查询帮助我获得了所需的确切输出

String query =
                "PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>" +
                        "PREFIX owl: <http://www.w3.org/2002/07/owl#> " +
                        "PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#> " +
                        "PREFIX ns0: <http://www.Jobsurveillance.com/CV#> " +
                        "SELECT (str(?object) as ?label) " +
                        "(str(?object2) as ?label2) " +
                        "(str (?phoneNumobj) as ?label3) " +
                        "(str (?degreeobj) as ?label4) " +
                        "(str (?jobcatObj) as ?label5) " +
                        "(str (?candidateSkill) as ?label8) " +
                        "(str (?jobSkill) as ?label7) WHERE { " +
                        " ?person ns0:Full_Name ?object ." +
                        " ?person ns0:Email_Id ?object2 ." +
                        " ?person ns0:Phone_Number ?phoneNumobj ." +
                        " ?person ns0:Degree ?degreeobj ." +
                        " ?person ns0:Job_Category ?jobcatObj ." +
                        " ?person ns0:Skills ?candidateSkill ." +
                        " ?job ns0:JobID  \"" + jobID + "\"." +
                        " ?job ns0:JobSkills ?jobSkill ." +
                        " filter (?jobSkill = ?candidateSkill)" +
                        "}";

在上面的代码中，以下行是

filter (?jobSkill = ?candidateSkill))

更正为

filter (?jobSkill = ?candidateSkill)

在SPARQL查询中使用多个过滤器

1 个答案: