How to return all S->P->O triples from a starting resource to a specified path depth?

My goal is to graphically represent the S->P->O relations within a depth two edges from the specified resource, p:Person_1. I want all relations within that path length to be returned from my query as ?s, ?p, ?o for further processing in my graphical application.

I tried the first query below which gives me my first set of ?s ?p ?o with repeats, then ?p2, ?o2, ?p3, ?o3 as additional columns in the result. I want to bind ?p2 and ?p3 to ?p, ?o2 and ?o3 to ?o.

  SELECT *
    WHERE {
        p:Person_1 ?p ?o .
        BIND("p:Person_1" as ?s)
        OPTIONAL{
            ?o ?p2 ?o2 .
        }
        OPTIONAL{
            ?o2 ?p3 ?o3 .
        }
    }

Then, based on How do I construct get the whole sub graph from a given resource in RDF Graph?, I tried using CONSTRUCT to return the graph.

PREFIX p: <http://www.example.org/person/> 
PREFIX x: <example.org/foo/>

construct { ?s ?p ?o }
FROM <http://localhost:8890/MYGRAPH>
where { p:Person_1 (x:|!x:)* ?s . 
        ?s ?p ?o . 
}

I am using Virtuoso and I get the error:

Virtuoso 37000 Error SP031: SPARQL compiler: Variable ?_::trans_subj_9_3 in T_IN list is not a value from some triple

I could post-process the result from my first query but I want to learn how to do this correctly with SPARQL, preferably on Virtuoso.

Update after testing the advice from @AKSW : Both CONSTRUCT and SELECT statements work with the pattern suggested.

CONSTRUCT { ?s ?p ?o }
FROM <http://localhost:8890/MYGRAPH>
where { p:Person_1 (x:foo|!x:bar)* ?s . 
        ?s ?p ?o . 
} LIMIT 100

and:

SELECT s ?p ?o 
FROM <http://localhost:8890/MYGRAPH>
where { p:Person_1 (x:foo|!x:bar)* ?s . 
        ?s ?p ?o . 
} LIMIT 100

The SELECT results in several duplicates that cannot be removed using DISTINCT, which results in an error that I assume is due to the 'datatype' of some of the returned values.

Virtuoso 22023 Error SR066: Unsupported case in CONVERT (DATETIME -> IRI_ID)

It appears some post-SPARQL processing is in order.

This gets me most of the way there. Still hoping I can find a solution for SPARQL that is like Cypher's "number of hops away" :

OPTIONAL MATCH path=s-[*1..3]-(o)

Here is a SPARQL query that works in Virtuoso. Note the SPARQL W3C standard does not support this syntax and it will fail in other triplestores.

PREFIX p: <http://www.example.org/person/> 
PREFIX x: <example.org/foo/>
# CONSTRUCT {?s ?p ?o}  # If you wish to return the graph
SELECT ?s ?p ?o   # To return the triples
FROM <http://localhost:8890/MYGRAPH>
where { p:Person_1 (x:foo|!x:bar){1,3} ?s  .
  ?s ?p ?o .
}LIMIT 100

See also K. Idehen's wiki entry here: http://linkedwiki.com/exampleView.php?ex_id=141

And thanks to @Joshua Taylor for advice in the same area.

Kingsley Uyi Idehen

Working Drafts of SPARQL 1.1 Property Paths included the {n,m} operator for handling this issue, which was implemented (and will remain supported) in Virtuoso. Here's a tweak to @tim's response.

Live SPARQL Query Results Page using the DBpedia endpoint (which is a Virtuoso instance).
Live SPARQL Query Definition Page that opens up query source code in the default DBpedia query editor.

Actual Query Example:

PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> 
SELECT DISTINCT ?s AS ?Entity 
                ?o AS ?Category 
WHERE { 
        ?s  rdf:type       <http://dbpedia.org/ontology/AcademicJournal> ; 
            rdf:type{1,3}  ?o 
      } 
LIMIT 100

Kingsley Uyi Idehen

Should you be looking for LinkedIn-like presentation of Contact Networks and Degrees of Separation between individuals, here is an example using Virtuoso-specific SPARQL Extensions that solve this particular issue:

SELECT ?o                                              AS ?WebID
       ((SELECT COUNT (*) WHERE {?o foaf:knows ?xx}))  AS ?contact_network_size
       ?dist                                           AS ?DegreeOfSeparation 
       <http://www.w3.org/People/Berners-Lee/card#i>   AS ?knowee
  WHERE
    {
      {
        SELECT ?s ?o
        WHERE
          {
            ?s foaf:knows ?o
          }
      } OPTION (TRANSITIVE, t_distinct, t_in(?s), t_out(?o), t_min (1), t_max (4), t_step ('step_no') AS ?dist) .
      FILTER (?s= <http://www.w3.org/People/Berners-Lee/card#i>)
      FILTER (isIRI(?s) and isIRI(?o))
    }
  ORDER BY ?dist DESC (?contact_network_size)
  LIMIT 500

Note: this approach is the only way (at the current time) to expose actual relational hops between entities in an Entity Relationship Graph that includes Transitive relations.

Live Link to Query Results
Live Link to Query Source Code

Kingsley Uyi Idehen

Bearing in mind that the r{n,m} operator was deprecated in the final SPARQL 1.1 (but will remain supported in Virtuoso), you can use r/r?/r? instead of r{1,3}, if you want to work strictly off the current spec:

PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> 
SELECT DISTINCT ?s AS ?Entity 
                ?o AS ?Category 
WHERE { 
        ?s  rdf:type                          <http://dbpedia.org/ontology/AcademicJournal> ; 
            rdf:type / rdf:type? / rdf:type?  ?o 
       }
LIMIT 100

Here's a live example, against the DBpedia instance hosted in Virtuoso.

来源：https://stackoverflow.com/questions/41025079/how-to-return-all-s-p-o-triples-from-a-starting-resource-to-a-specified-path-d

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