Question

我有两个SPARQL查询，一个获取总数没有缺陷，第二个获取总数没有。测试用例。我想将一个计数结果除以另一个计数，即（缺陷总数/测试用例总数）。如何使用除法运算符划分两个查询结果。

任何人都可以建议我同意吗？

    PREFIX process: <http://jazz.net/ns/process#>
    PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
    PREFIX merge: <http://jazz.net/ns/lqe/merge/>
    PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
    PREFIX owl: <http://www.w3.org/2002/07/owl#>
    PREFIX oslc: <http://open-services.net/ns/core#>
    PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
    PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
    PREFIX rtc_merged: <http://jazz.net/ns/lqe/merge/gensym/cm/>
    PREFIX oslc_config: <http://open-services.net/ns/config#>
    PREFIX dcterms: <http://purl.org/dc/terms/>


    SELECT DISTINCT
      ?rtc_merged_Defect1_projectArea
      ?rtc_merged_Defect1_shortId_COUNT
    WHERE{
      {
        SELECT DISTINCT
          ?rtc_merged_Defect1_projectArea
          (COUNT(DISTINCT ?rtc_merged_Defect1_shortId) AS ?rtc_merged_Defect1_shortId_COUNT)
        WHERE{
          VALUES( ?rtc_merged_Defect1_paUri )
          {
            ( <https://localhost:9443/ccm/process/project-areas/__ANzQBasEeegkPxZTDywOA> )
          }
          ?rtc_merged_Defect1_uri process:projectArea ?rtc_merged_Defect1_paUri.
          ?rtc_merged_Defect1_uri oslc:instanceShape ?rtc_merged_Defect1_uri_instanceShape.
          ?rtc_merged_Defect1_uri_instanceShape merge:mergeShape rtc_merged:Defect.
          OPTIONAL {
            ?rtc_merged_Defect1_uri process:projectArea ?rtc_merged_Defect1_projectArea_enumValue.
            OPTIONAL {?rtc_merged_Defect1_projectArea_enumValue rdfs:label|dcterms:title ?rtc_merged_Defect1_projectArea_enumName.}
          }
          OPTIONAL {?rtc_merged_Defect1_ver dcterms:isVersionOf ?rtc_merged_Defect1_uri; rdf:type oslc_config:VersionResource.}
          ?rtc_merged_Defect1_uri oslc:shortId ?rtc_merged_Defect1_shortId.
          BIND( IF (bound(?rtc_merged_Defect1_ver), concat(str(?rtc_merged_Defect1_uri), "?oslc_config.context="), ?rtc_merged_Defect1_uri) as ?rtc_merged_Defect1)
          BIND( IF(bound(?rtc_merged_Defect1_projectArea_enumName), ?rtc_merged_Defect1_projectArea_enumName, IF(bound(?rtc_merged_Defect1_projectArea_enumValue), ?rtc_merged_Defect1_projectArea_enumValue, "")) as ?rtc_merged_Defect1_projectArea)
        }
        GROUP BY ?rtc_merged_Defect1_projectArea
      }
    }  //This is my first query.
===============================================================


   PREFIX process: <http://jazz.net/ns/process#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX merge: <http://jazz.net/ns/lqe/merge/>
PREFIX xsd: <http://www.w3.org/2001/XMLSchema#>
PREFIX owl: <http://www.w3.org/2002/07/owl#>
PREFIX oslc: <http://open-services.net/ns/core#>
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rqm_merged: <http://jazz.net/ns/lqe/merge/gensym/qm/>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
PREFIX oslc_config: <http://open-services.net/ns/config#>
PREFIX dcterms: <http://purl.org/dc/terms/>


SELECT DISTINCT
  ?rqm_merged_QMTestCase1_projectArea
  ?rqm_merged_QMTestCase1_shortId_COUNT
WHERE{
  {
    SELECT DISTINCT
      ?rqm_merged_QMTestCase1_projectArea
      (COUNT(DISTINCT ?rqm_merged_QMTestCase1_shortId) AS ?rqm_merged_QMTestCase1_shortId_COUNT)
    WHERE{
      VALUES( ?rqm_merged_QMTestCase1_paUri )
      {
        ( <https://localhost:9443/qm/process/project-areas/_EazMoBatEeeR-7d5-ZtSzw> )
      }
      ?rqm_merged_QMTestCase1_uri process:projectArea ?rqm_merged_QMTestCase1_paUri.
      ?rqm_merged_QMTestCase1_uri oslc:instanceShape ?rqm_merged_QMTestCase1_uri_instanceShape.
      ?rqm_merged_QMTestCase1_uri_instanceShape merge:mergeShape rqm_merged:QM%20Test%20Case.
      OPTIONAL {
        ?rqm_merged_QMTestCase1_uri process:projectArea ?rqm_merged_QMTestCase1_projectArea_enumValue.
        OPTIONAL {?rqm_merged_QMTestCase1_projectArea_enumValue rdfs:label|dcterms:title ?rqm_merged_QMTestCase1_projectArea_enumName.}
      }
      OPTIONAL {?rqm_merged_QMTestCase1_ver dcterms:isVersionOf ?rqm_merged_QMTestCase1_uri; rdf:type oslc_config:VersionResource.}
      ?rqm_merged_QMTestCase1_uri oslc:shortId ?rqm_merged_QMTestCase1_shortId.
      BIND( IF (bound(?rqm_merged_QMTestCase1_ver), concat(str(?rqm_merged_QMTestCase1_uri), "?oslc_config.context="), ?rqm_merged_QMTestCase1_uri) as ?rqm_merged_QMTestCase1)
      BIND( IF(bound(?rqm_merged_QMTestCase1_projectArea_enumName), ?rqm_merged_QMTestCase1_projectArea_enumName, IF(bound(?rqm_merged_QMTestCase1_projectArea_enumValue), ?rqm_merged_QMTestCase1_projectArea_enumValue, "")) as ?rqm_merged_QMTestCase1_projectArea)
    }
    GROUP BY ?rqm_merged_QMTestCase1_projectArea
  }
}

Answer 1

解决方案是使用两个子SELECT查询（我在这里使用存根，因为您当前的SPARQL查询令人困惑（重复并使用不必要的子SELECT））：

SELECT (?cnt1/?cnt2 AS ?result) WHERE {

// first query
{
  SELECT (COUNT(?s) AS ?cnt1) WHERE {
     ...
  }
}

// second query
{
  SELECT (COUNT(?s) AS ?cnt2) WHERE {
     ...
  }
}

}

SPARQL从两个查询中划分两个值

1 个答案: