I am trying to run the following SPARQL query on the public endpoint of DBPedia:
select distinct ?o ?olabel where {
{ select distinct ?s where {
?s <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>/<http://www.w3.org/2000/01/rdf-schema#subClassOf>{,6} <http://dbpedia.org/ontology/FictionalCharacter>
. ?s <http://dbpedia.org/property/creator> ?oo } limit 100000 }
. ?s <http://dbpedia.org/property/creator> ?o
MINUS { ?o <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>/<http://www.w3.org/2000/01/rdf-schema#subClassOf>{,6} <http://dbpedia.org/class/yago/Communicator109610660>}
# MINUS { ?o <http://www.w3.org/1999/02/22-rdf-syntax-ns#type>/<http://www.w3.org/2000/01/rdf-schema#subClassOf>{,6} <http://dbpedia.org/class/yago/Creator109614315>}
. ?o <http://www.w3.org/2000/01/rdf-schema#label> ?olabel . FILTER (lang(?olabel)=''||lang(?olabel)='en')
} LIMIT 100
It works ok.
But then I uncomment the commented line.
And then the query fails.
From what I understand, multiple MINUS clauses are authorized in SPARQL.
Can anyone give me pieces of advice to rewrite my query so the multiple MINUS clauses are taken into account by DBPedia?
One of the nice things about standards is that we can look things up in the specifications, so after looking at productions 66, 56, and 54 at https://www.w3.org/TR/sparql11-query/#grammar I can confirm that multiple MINUS clauses are allowed by the standard.
I want to query all triples with a certain predicate p. The query should only return triples (s,p,o) where the other direction (o,p,s) does not exist.
How can I make such a query?
That's pretty easy:
SELECT ?s ?p ?o {
?s ?p ?o
MINUS { ?o ?p ?s }
}
FILTER NOT EXISTS instead of MINUS would also work. Replace ?p in the query with the desired predicate, or use something like FILTER (?p=ex:myPredicate) if you want ?p in the result.
I have a graph created by sponging web pages. I'm trying to write a query to exclude all triples where the subject is a web page. I can get these triples simply by
FILTER (STRENDS((STR(?i)), "html"))
but what's the best way to exclude them. Basically I need all {?s ?p ?o} minus the subset {?i ?p ?o}.
This
SELECT *
{ ?s ?p ?o
FILTER ( ! STRENDS( STR(?s), "html") )
}
filters the outcome the pattern.
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.
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
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
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.
Assuming we have a dataset including the following 3 RDF triples. I want to retrieve all these three triples using sparql query.
http://test.com/11/META_C1-INST predicate1 http://test.com/NO_CONTEXT/META_C0059714-INST
http://test.com/23/META_C1-INST predicate1 http://test.com/NO_CONTEXT/META_C0059714-INST
http://test.com/43/META_C1-INST predicate1 http://test.com/NO_CONTEXT/META_C0142817-INST
Here is the query I wrote but I know it is not right:
prefix pred:<http://test.com/>
select distinct *
where {pred:META_C1-INST ?p ?o};
How should I say: "Look for any triples with the subject from http://test.com/??/ where ?? can be anything like 11, 23, S23?
You can transform the URI into a string and then filter based on what it starts with. The keyword you need to use is STRSTARTS:
select distinct *
where{
?s ?p ?o
filter(STRSTARTS(str(?s), "http://test.com/"))
}
or you can use regular expressions with the ^ sign:
select distinct *
where{
?s ?p ?o
FILTER regex(str(?s), "^http://test.com/") .
}