Return a nested data structure from a SPARQL query

前端 未结 3 851
不知归路
不知归路 2021-02-15 16:22

If I have a graph with this kind of structure:

@prefix  :        .
@prefix  rdf:     .
         


        
相关标签:
3条回答
  • 2021-02-15 17:08

    Both answers do not mention the danger of Cartesian explosion. If you have mulivalued fields of the root person, these will be multiplied by the multiple related persons, creating unnecessary rows. A construct may collapse those extra rows, but still: the worry should not make them in the first place. So use Union instead of straight patterns.

    0 讨论(0)
  • 2021-02-15 17:10

    You could use a CONSTRUCT query with JSON-LD Framing.

    Example query (on DBpedia endpoint)

    CONSTRUCT
    {
      ?person  rdf:type       foaf:Person ;
               dbo:birthName  ?name1s ;
               dbo:birthDate  ?date1s ;
               dbo:spouse     ?spouse .
      ?spouse  rdf:type       foaf:Person ; 
               dbo:birthName  ?name2s ;
               dbo:birthDate  ?date2s .
    }
    WHERE
    {
      ?person  dbo:birthName  ?name1 ;
               dbo:birthDate  ?date1 ;
               dbo:spouse     ?spouse .
      ?spouse  dbo:birthName  ?name2 ;
               dbo:birthDate  ?date2 .
      BIND (str(?name1) AS ?name1s)
      BIND (str(?date1) AS ?date1s)
      BIND (str(?name2) AS ?name2s)
      BIND (str(?date2) AS ?date2s)
      VALUES (?person) { ( dbr:Brad_Pitt ) }
    }
    

    Output (in JSON-LD format with context)

    { "@context": {
        "spouse": { "@id": "http://dbpedia.org/ontology/spouse"},
        "birthDate": { "@id": "http://dbpedia.org/ontology/birthDate" },
        "birthName": { "@id": "http://dbpedia.org/ontology/birthName" } },
      "@graph": [
        { "@id": "http://dbpedia.org/resource/Angelina_Jolie",
          "birthName": "Angelina Jolie Voight",
          "birthDate": "1975-06-04" },
        { "@id": "http://dbpedia.org/resource/Brad_Pitt",
          "@type": "http://xmlns.com/foaf/0.1/Person",
          "birthName": "William Bradley Pitt",
          "spouse": [ "http://dbpedia.org/resource/Angelina_Jolie",
                      "http://dbpedia.org/resource/Jennifer_Aniston" ],
          "birthDate": "1963-12-18" },
        { "@id": "http://dbpedia.org/resource/Jennifer_Aniston",
          "birthName": "Jennifer Joanna Aniston",
          "birthDate": "1969-02-11" }
    ] }
    

    JSON-LD Frame (very simple)

    {
      "@context": {"dbo": "http://dbpedia.org/ontology/",
                   "dbr": "http://dbpedia.org/resource/",
                   "foaf": "http://xmlns.com/foaf/0.1/"},
      "dbo:spouse": {
       }
    }
    

    Framed JSON-LD (playground)

    {
      "@context": {
        "dbo": "http://dbpedia.org/ontology/",
        "dbr": "http://dbpedia.org/resource/",
        "foaf": "http://xmlns.com/foaf/0.1/"
      },
      "@graph": [
        {
          "@id": "dbr:Brad_Pitt",
          "@type": "foaf:Person",
          "dbo:birthDate": "1963-12-18",
          "dbo:birthName": "William Bradley Pitt",
          "dbo:spouse": [
            {
              "@id": "dbr:Angelina_Jolie",
              "@type": "foaf:Person",
              "dbo:birthDate": "1975-06-04",
              "dbo:birthName": "Angelina Jolie Voight"
            },
            {
              "@id": "dbr:Jennifer_Aniston",
              "@type": "foaf:Person",
              "dbo:birthDate": "1969-02-11",
              "dbo:birthName": "Jennifer Joanna Aniston"
            }
          ]
        }
      ]
    }
    

    Some discussion

    JSON-LD Framing is an unofficial, yet well implemented specification that describes a deterministic layout for serializing an RDF graph into a particular JSON-LD document layout.

    Obviously, with blank nodes property lists, one can achieve something structurally similar to the output you want:

    Brad_Pitt
            dbo:birthName   "William Bradley Pitt" ;
            dbo:birthDate   "1963-12-18" .
            dbo:spouse  [   dbo:birthName   "Angelina Jolie Voight" ;
                            dbo:birthDate   "1975-06-04" ] ,
                        [   dbo:birthName   "Jennifer Joanna Aniston" ;
                            dbo:birthDate   "1969-02-11" ] .
    

    However, this is Turtle, not JSON, and nobody can garantee that these blank nodes property lists will be used in serialization.

    0 讨论(0)
  • 2021-02-15 17:22

    You're conflating the query result itself (essentially an abstract table structure) with the syntax in which that result is written (in your case, a customized nested JSON structure).

    Don't try to do tricks with group concatenation in this case. Just do this query:

    SELECT ?given ?family ?friend_given ?friend_family
    WHERE {
      ?person foaf:givenName ?given ;
              foaf:familyName ?family .
      ?person :knows ?friend .
      ?friend foaf:givenName ?friend_given ;
              foaf:familyName ?friend_family .
    }
    GROUP BY ?family ?given
    

    Which results in a result like this:

    given  family  friend_given friend_family
    -------------------------------------------- 
    Alice  Lidell  Bob          Doe
    Alice  Lidell  Hwa          Choi
    

    And then let a custom streaming result writer write the result to the nested syntax format you require. Given that the query groups by name, the writer can safely assume that subsequent rows with the same given and family names "belong together".

    Alternatively, use a CONSTRUCT query instead of a SELECT, and post-process the retrieved RDF graph (which accurately represents the tree structure you're after).

    0 讨论(0)
提交回复
热议问题