Is it possible to execute SPARQL construct while adding information outside the scope of query? e.g., I want to execute SPARQL construct while defining enumeration information like this:
PREFIX skos:<http://www.w3.org/2004/02/skos/core#>
construct {
?s a skos:Concept
?s ex:index <enumeration starting from 1 -- this is just a sample>
}
where {
?s a skos:Concept
}
is it possible to do something like that with pure SPARQL? what are the alternatives?
* Additional Information *
Probably I am not explained my problem clearly, so basically I want to achieve the following (assuming that ex:index is a valid datatypeProperty):
== Initial RDF triples ==
#prefix skos:<http://www.w3.org/2004/02/skos/core#>
#prefix ex: <http://example.org/> .
ex:abc rdf:type skos:Concept .
ex:def rdf:type skos:Concept .
...
ex:endOfSample rdf:type skos:Concept .
== RDF triples after SPARQL Update execution ==
#prefix skos:<http://www.w3.org/2004/02/skos/core#>
#prefix ex: <http://example.org/> .
ex:abc rdf:type skos:Concept ;
ex:index 1 .
ex:def rdf:type skos:Concept ;
ex:index 2 .
...
ex:endOfSample rdf:type skos:Concept ;
ex:index <endOfSampleNumber> .
You can construct any valid RDF value in a CONSTRUCT. However the query will fail if any of the variables in the CONSTRUCT graph pattern is unbound after executing the WHERE graph. I.e. there can be no binding for ?p in your query and the CONSTRUCT will never execute.
This is an example that should get you started:
PREFIX skos:<http://www.w3.org/2004/02/skos/core#>
PREFIX ex:<http://example.org/construct#>
construct {
ex:someProp a owl:ObjectProperty .
?s ex:someProp (1 2 3)
}
where {
?s a skos:Concept
}
This will result in the construction of seven triples for the property value and the list structure.
The ex:someProp is added because there isn't a good object property in SKOS for ad-hoc lists. It would be best to define the property with some semantic meaning. Also note that while the {ex:someProp a owl:ObjectProperty} triple will be asserted for each match of {?s a skos:Concept}, it is the same triple, hence there will be only one in the end. The price is efficiency, so asserting the property outside of this query would be a better choice - it is included in the above query for the sake of example completeness.
Related
I am in a learning phase of SPARQL and ontology building. I have a model and I would like to add a new concept class to multiple concepts in a model using regex/filter.
I have following concepts:
A647674
A878678
RR36868
DD36868
The expected output is :
A647674
A878678
RR36868 rdf:type http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule
DD36868 rdf:type http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule
I am using below SPARQL query to do this.
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
INSERT {
?s rdf:type 'http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule' .
}
WHERE
{
{?s ?p ?o .
filter regex(str(?s), "http://ontology.aaaaaaa.com/drugs/aaaaaaa#RR-").
}
union
{?s ?p ?o .
filter regex(str(?s), "http://ontology.aaaaaaa.com/drugs/aaaaaaa#DD").
}
};
#LIMIT 100
I am getting below error using above query.
OmServerGenericException[message="http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule",responseCode=500]
Caused by: org.apache.jena.rdf.model.ResourceRequiredException: "http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule"
Any help is highly appreciated
You are providing a string value:
?s rdf:type 'http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule' .
You need to provide a URI value:
?s rdf:type <http://schemas.aaaaaaa.com/ontologies/drug#SmallMolecule> .
Or you could define a prefix (PREFIX drug: <http://schemas.aaaaaaa.com/ontologies/drug#>) to use it like this:
?s rdf:type drug:SmallMolecule .
(The suggestions given in my answer to your previous question apply here, too: you could use STRSTARTS instead of REGEX, and one FILTER with || instead of UNION.)
Imagine you do something crazy and store your object-oriented model as an RDF graph.
shows a simplified example of the inheritance hierarchy and the associated attributes.
In practice, you get such graph structure if you translate some UML class diagram into RDFS.
The question is: what SPARQL query can deliver all the predicate-object pairs necessary to instantiate a particular resource of "Class C". In other words: how do you get all the predicate-object pairs along the whole inheritance chain (only single inheritance).
Given this diagram, the predicate-object pairs of all members of the class:ClassC is simply:
SELECT ?inst ?p ?o
WHERE {
?inst a :ClassC .
Inst ?p ?o .
Keep in mind that there is not property inheritance in RDF/RDFS. If you want to find all of property/values pairs for ClassA with entailments for subclasses then useL
SELECT ?inst ?p ?o
WHERE {
?cls rdfs:subClassOf* :ClassA .
?inst a ?cls .
?inst ?p ?o
}
In this respect, RDFS works a bit backwards of one's expectations of OO inheritance.
With the info from #scotthenninger the following query did the job:
SELECT ?p ?o
WHERE {
:ClassC rdfs:subClassOf* ?anySuperClass .
?anySuperClass ?p ?o .
}
edit:
Similar query gets all the self-defined properties and their range along the inheritance chain:
SELECT ?prop ?obj
WHERE {
:ClassC rdfs:subClassOf* ?anySuperClass .
?prop rdfs:domain ?anySuperClass .
?prop rdfs:range ?obj .
}
End results combined:
foo:ID xsd:string
foo:name xsd:string
rdfs:comment xsd:string
foo:similarTo :ClassD
I have create an endpoint SPAQL on OpenLink Virtuoso.
All work well, but i have to access on the data in a Container, in particular a rdf:Seq.
I have a Seq like this:
<myrdf:has_stoptimes>
<rdf:Seq rdf:about="http://test.com/343">
<rdf:li>
<myrdf:StopTime rdf:about="http://test.com/StopTime/434">
...
</ns0:StopTime>
</rdf:li>
<rdf:li>
<myrdf:StopTime rdf:about="http://test.com/StopTime/435">
...
</ns0:StopTime>
</rdf:li>
</rdf:Seq>
Now i see that to access data in a container i can use rdfs:member or FILTER (strstarts(str(?prop), str(rdf:_)) how is explained here
But for my project i have to adopt the first solution because i'm working with Silk and i will use the code syntax like ?a/myrdf:has_stoptimes/rdfs:member without use of "complex" filter.
I have tried to follow this guide but querying the endpoint nothing work how i hoped.
So my question is: how can i query ?a/myrdf:has_stoptimes/rdfs:member on a Virtuoso endpoint SPARQL?Which inference rule i have to add in endpoint SPARQL?
Thank you in advance
UPDATE
I have created the following inference rules in Virtuoso:
ttlp (' #prefix rdfs: .
#prefix rdf: .
rdfs:Container rdf:type rdfs:Class ; rdfs:subClassOf rdfs:Resource .
rdfs:ContainerMembershipProperty a rdfs:Class ; rdfs:subClassOf rdf:Property .
rdf:Seq rdf:type rdfs:Class ; rdfs:subClassOf rdfs:Container .
rdfs:member rdf:type rdf:Property ; rdfs:domain rdfs:Resource ; rdfs:range rdfs:Resource .
', '', 'http://localhost:8890/schema/test') ;
Nothing work querying the SPARQL endpoint like:
define input:inference "http://localhost:8890/schema/property_rules1"
SELECT *
FROM
WHERE {?sep a rdf:Seq.
?seq rdfs:member ?p}
After i tried adding the follow line to the ttl file: rdf:_1 rdfs:subPropertyOf rdfs:member . In this way it work but obviously the results are only for the first element of the container. So is very unconvenient add a line for all of rdf:_n, and i think this is only a temporary solution, it is not correct.
I have tried to add an RDF dump on SILK 2.6.1, and on the section SPARQL of the data source if i run the query:
SELECT *
FROM
WHERE {?sep a rdf:Seq.
?seq rdfs:member ?p}
I obtain the correct result, without specify any inference rules. So i think that in this functionality of SILK there is something that i’m missing in my endpoint SPARQL or am i saying nonsense things?
You can't use variables in property paths, so you can't actually do
?x ?a/has_stoptimes/rdfs:member ?y
Instead, you have to use another variable or blank node in between:
?x ?a ?z . ?z has_stoptimes/rdfs:member ?y
?x ?a [ has_stoptimes/rdfs:member ?y ] .
I need a Sparql query to recover the Type of a specific DBpedia resource. Eg.:
pt.DBpedia resource: http://pt.dbpedia.org/resource/Argentina
Expected type: Country (as can be seen at http://pt.dbpedia.org/page/Argentina)
Using pt.DBpedia Sparql Virtuoso Interface (http://pt.dbpedia.org/sparql) I have the query below:
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
select ?l ?t where {
?l rdfs:label "Argentina"#pt .
?l rdf:type ?t .
}
But it is not recovering anything, just print the variable names. The virtuoso answer.
Actually I do not need to recover the label (?l) too.
Anyone can fix it, or help me to define the correct query?
http in graph name
I'm not sure how you generated your query string, but when I copy and paste your query into the endpoint and run it, I get results, and the resulting URL looks like:
http://pt.dbpedia.org/sparql?default-graph-uri=http%3A%2F%2Fpt.dbpedia.org&sho...
However, the link in your question is:
http://pt.dbpedia.org/sparql?default-graph-uri=pt.dbpedia.org%2F&should-sponge...
If you look carefully, you'll see that the default-graph-uri parameters are different:
yours: pt.dbpedia.org%2F
mine: http%3A%2F%2Fpt.dbpedia.org
I'm not sure how you got a URL like the one you did, but it's not right; the default-graph-uri needs to be http://pt.dbpedia.org, not pt.dbpedia.org/.
The query is fine
When I run the query you've provided at the endpoint you've linked to, I get the results that I'd expect. It's worth noting that the label here is the literal "Argentina"#pt, and that what you've called ?l is the individual, not the label. The individual ?l has the label "Argentina"#pt.
We can simplify your query a bit, using ?i instead of ?l (to suggest individual):
select ?i ?type where {
?i rdfs:label "Argentina"#pt ;
a ?type .
}
When I run this at the Portuguese endpoint, I get these results:
If you don't want the individual in the results, you don't have to select it:
select ?type where {
?i rdfs:label "Argentina"#pt ;
a ?type .
}
or even:
select ?type where {
[ rdfs:label "Argentina"#pt ; a ?type ]
}
If you know the identifier of the resource, and don't need to retrieve it by using its label, you can even just do:
select ?type where {
dbpedia-pt:Argentina a ?type
}
type
==========================================
http://www.w3.org/2002/07/owl#Thing
http://www.opengis.net/gml/_Feature
http://dbpedia.org/ontology/Place
http://dbpedia.org/ontology/PopulatedPlace
http://dbpedia.org/ontology/Country
http://schema.org/Place
http://schema.org/Country
I have an ontology where arc_cfp is an individual of class Arc. I would like to know how could I get all the data properties of the individual, given that I have the individual's URI?
Basically, I am doing this:
SELECT ?idRef ?name ?src ?dst ?perf
WHERE
{
?x rdf:type http://www.semanticweb.org/ontologies/2012/1/graph.owl#arc_cfp .
?x graph:idRef_arc ?idRef .
?x graph:name_arc ?name .
?x graph:hasSource ?src .
?x graph:hasDestination ?dst .
?x graph:hasPerformatif ?perf .
}
I am pretty sure, using rdf:type is the problem. But, I have no idea what I need to use.
Thanks.
~Codera
Assuming you want a purely exploratory query of the form "give me all the triples about a subject" it should look the following:
SELECT *
WHERE
{
<http://example.org/SomeThing> ?p ?o
}
This will give you all predicate object pairs associated with the constant URI you pass in. If you are interesting in incoming as well as outgoing properties you could do the following instead:
SELECT *
WHERE
{
{ <http://example.org/SomeThing> ?p ?o }
UNION
{ ?s ?p <http://example.org/SomeThing> }
}
You can also use a DESCRIBE query to grab all the RDF data about a Resource.
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
DESCRIBE ?x
WHERE
{
?x rdf:type http://www.semanticweb.org/ontologies/2012/1/graph.owl#arc_cfp .
}
P.S. Don't forget to put prefixes in your queries.