https://graphdb.ontotext.com/documentation/standard/exploring-data.html#create-your-own-visual-graph describes the ability to create a visual graph using SPARQL queries, and for the "Edge Basics" step it says:
Edge basics: This query SELECT the ?label binding that determines the text of the edge. If empty, the edge IRI’s local name is used.
In the UI for this step, there is also the "RDFS or SKOS label" sample query, which fills:
# Note that ?edge is the edge's IRI and must be used in the query.
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX skos: <http://www.w3.org/2004/02/skos/core#>
SELECT ?label {
# Select either rdfs:label or skos:prefLabel as the label
?edge rdfs:label | skos:prefLabel ?label .
}
My question is how do I leverage this capability? More specifically, what is the edge IRI (i.e. ?edge) that is the subject of the rdfs:label statement?
Is it simply the RDF predicate in the graph, as in the following?
:someSubject :somePredicate :someObject .
:somePredicate rdfs:label "Some friendly label for every :somePredicate relation" .
Is it possible somehow to use RDF-star, as in the following?
:someSubject :somePredicate :someObject .
<<:someSubject :somePredicate :someObject>> rdfs:label "A friendly label for the particular relation between :someSubject and :someObject" .
It appears to be the former case, as I get the following result when I load the above statements and use the sample query above:
someSubject --"Some friendly label for every :somePredicate relation"--> someObject
I'm curious if there is any way to leverage RDF-star statements in order to render edge-specific labels?
Related
My SPARQL code out of the Learning SPARQL book:
Contruct
{
?s dm:problem dm:prob29 .
dm:prob29 rdfs:label "Location value must be a URI." .
}
WHERE
{
?s dm:location ?city .
FILTER (!(isURI(?city)))
}
-- creates a file like this:
dm:prob29 rdfs:label "Location value must be a URI." .
d:item693 dm:problem dm:prob29 .enter
Why does he create the "Location value must be URI" triple first, when in the Contruct command dm:prob29 etc. is shown first? I am not really sure how this work?
The order of such triples is arbitrary, and has no importance nor meaning, in the context of a CONSTRUCT query's output.
I have the following query
CONSTRUCT{
?entity a something;
a label ?label .
}
WHERE
{
?entity a something;
a label ?label .
BIND(CONCAT(STR( ?label ), " | SOME ADDITIONAL TEXT I WOULD LIKE TO APPEND MANUALLY") ) AS ?label ) .
}
I simply want to concatenate some text with ?label, however when running the query I get the following error:
BIND clause alias '?label' was previously used
I only want to return a single instance of ?label hence, I defined it in the construct clause.
The error message seems to be accurate, but is only the first of many you will get with this query. The usual request to take a look at some SPARQL learning resources to at least understand the basics of triple-based graph pattern matching, along with, a couple of hints one what to look for. CONSTRUCT isn't a bad place to start, and the following should almost do what I think you intend:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
CONSTRUCT{
?entity rdfs:label ?label .
}
WHERE
{
?entity a ex:something ;
rdfs:label ?oldlabel .
BIND(CONCAT(STR( ?oldlabel ), " | SOME ADDITIONAL TEXT I WOULD LIKE TO APPEND MANUALLY") ) AS ?label ) .
}
There's quite a few things different about that query, so take a look to see if it accurately does what you want. One hint is the syntactic difference between using '.' and ';' to separate the triple patterns. Another is that each clause defines either a URL, using a qname in the example, or a variable, prefixed by a '?'. Neither 'label' or 'something' are valid.
I say "almost" because CONSTRUCT only returns a set of triples. To modify the labels, which I think is the intent, you need to use SPARQL Update, i.e.:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX ex: <http://example.org/example#>
DELETE {
?entity rdfs:label ?oldlabel .
}
INSERT{
?entity rdfs:label ?label .
}
WHERE
{
?entity a ex:something .
?entity rdfs:label ?oldlabel .
BIND(CONCAT(STR( ?oldlabel ), " | SOME ADDITIONAL TEXT I WOULD LIKE TO APPEND MANUALLY") AS ?label ) .
}
Note how the triple pattern finds matches for ?oldlabel and deletes them, inserting the newly bound ?label instead. This query assumes a default graph is defined that holds both the original data and the target for updates. If not then the graph needs to be specified using WITH or GRAPH. (Also included another hint on the syntactic difference between using '.' and ';' to separate triple patterns.)
I am trying this query:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
SELECT ?label
WHERE
{
?AGE rdfs:label ?label.
}
I need all the values of AGE from my model but instead this query is giving me other resources values which have the same property label .
For example I have connected the resource gender to have a property rdfs:label. So in my result I get both age values and gender values.
Can anybody tell me where am I wrong ?
It seems you may be assigning some semantics to the variable '?AGE'. SPARQL is a graph pattern matching language and anything with a '?' as the first character is a variable - or better yet, an unknown in the graph pattern match. I.e., the following is an equivalent query to yours:
SPARQL ?label
WHERE
{ ?s rdfs:label ?label .
}
This will find all triples that have a rdfs:label property and select the value of ?label.
If you have a specific resource you want to query, then specify that resource in the subject, for example:
PREFIX ex: <http://example.org/ex>
SPARQL ?label
WHERE
{ ex:AGE rdfs:label ?label .
}
So understanding the difference between an unknown (denoted by '?' (or '$')) and a known (a qname or a full URI) is important to understand how SPARQL performs graph pattern matching.
Lots of SPARQL learning material on the Web, so a suggestion is to look into some of these to learn some basics.
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 am doing some experiments with importing triples formulated in the turtle language
within the openrdf-workbench webapp in Tomcat, which has incorporated a SPARQL endpoint.
I wonder if with turtle, or, generally, in RDF / RDFS is it possible to add a certain predicate/object declaration on all (implicit) subjects conditionally to the existence of another predicate/object.
For example, if I have the following triples defined:
foo:a foo:b foo:c
foo:d foo:b foo:c
foo:e foo:b foo:c
foo:f foo:b foo:c
I would like to automatically add the following predicate/subject to all subjects that match predicate=foo:b and object=foo:c:
(implicit subject) foo:g foo:h
in order to automatically produce the following triples:
foo:a foo:g foo:h
foo:d foo:g foo:h
foo:e foo:g foo:h
foo:f foo:g foo:h
Is this possible?
Alternatively: is there any way to define some triples in order to enable SPARQL to find foo:a/d/e/f when queried for subjects that have foo:g foo:h as predicate/object?
Part 1 - Creating additional information
The first part of your question can be solved in one of two ways:
Using Inference
Using SPARQL Update
Inferencing
Inference is a technique whereby you define rules that infer additional triple based on your existing triples. You typically either use a pre-defined set of rules or use your own custom rules. I think Sesame only supports pre-defined rule sets out of the box so you may want to take a look at OWLIM which is an alternative back end that can be used with Sesame and has much more customisable rules AFAIK.
Inferencing can typically be applied in two ways, one where you only store the rules and you compute the additional information every time a rule fires and another where you pre-compute all the additional information and add it to your database. Which you will want to use depends on how you intend to use your system and there are performance trade offs involved. I'm not going into detail because that's really a whole other question - see Forward vs Backward Chaining for some discussion
SPARQL Update
Alternatively if your rules are relatively simple and you are OK with pre-computing the extra information and adding it to your database you can write SPARQL Updates to do this e.g.
PREFIX foo: <http://example.org/foo#>
INSERT
{
?x foo:g foo:h .
}
WHERE
{
?x foo:b foo:c .
}
Part 2 - Querying the Data
I am guessing you are fairly new to SPARQL because from what you've described this sounds trivial to me.
If I wanted to find all subjects which had the predicate foo:g and the object foo:h I would simply write the following:
PREFIX foo: <http://example.org/foo#>
SELECT ?x
WHERE
{
?x foo:g foo:h .
}
You can do this type of inference using OWL with an axiom of the form
p value a ⊑ q value b
which says that if something has a as a value for property p, then it also has b as a value for property q. As an example, here's an ontology with four individuals (a, b, c, d), two object properties (p, q), and the axiom (p value c ⊑ q value d).
#prefix : <http://example.org/add-predicate-object#> .
#prefix rdfs: <http://www.w3.org/2000/01/rdf-schema#> .
#prefix owl: <http://www.w3.org/2002/07/owl#> .
#prefix xsd: <http://www.w3.org/2001/XMLSchema#> .
#prefix rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#> .
<http://example.org/add-predicate-object> a owl:Ontology .
:p a owl:ObjectProperty .
:q a owl:ObjectProperty .
[ a owl:Restriction ;
owl:onProperty :p ;
owl:hasValue :c ;
rdfs:subClassOf [ a owl:Restriction ;
owl:onProperty :q ;
owl:hasValue :d ] . ] .
:a a owl:Thing, owl:NamedIndividual ; :p :c .
:b a owl:Thing, owl:NamedIndividual ; :p :c .
:c a owl:Thing, owl:NamedIndividual .
:d a owl:Thing, owl:NamedIndividual .
In Protégé, the axiom looks like this:
You can enable a reasoner and query for instances of q value d and see:
or you can browse to individuals and see the results: