I need to find all triples on DBpedia where http://dbpedia.org/resource/Benin is a subject or object. This query gives me the output that I want in a format that works the best for me (just three variables and no blank spaces):
PREFIX : <http://dbpedia.org/resource/>
SELECT * WHERE {
?s ?p ?o
FILTER (?s=:Benin OR ?o=:Benin)
}
I get similar results if I have this query:
PREFIX : <http://dbpedia.org/resource/>
SELECT * WHERE {
{:Benin ?p ?o}
UNION
{?s ?p :Benin}
}
However, the formatting of the latter is off. It first gives me p and o output leaving s blank and then s and p leaving o blank. Also, the first query takes more time to execute. I will be grateful for an explanation of the mechanics of how the two queries work and why there is a difference in the output.
However, the formatting of the latter is off
That's because both queries have different result sets together with SELECT *. The union joins the tuples, but since some tuples are missing parts, you get skewed output.
You can resolve the problem by explicitly listing and selecting the variables:
PREFIX : <http://dbpedia.org/resource/>
SELECT ?s ?p ?o WHERE {
{
?s ?p ?o
FILTER (?s=:Benin)
}
UNION
{
?s ?p ?o .
FILTER (?o=:Benin)
}
}
Note that this is still much faster on dbpedia than the OR filter.
The union will return duplicates when a tuple matches both filter expressions (i.e. :Benin ?p :Benin).
SELECT DISTINCT would remedy that at additional cost and since it looks like the problem is non-existent, I omitted it for improved performance.
Also, the first query takes more time to execute.
That's hard to say without the result of an EXPLAIN(), but my first guess would be that the equality filter is using the index, while the OR filter is using a full table scan. Virtuoso does not seem to generate good query plans for nested filters.
Try this --
PREFIX : <http://dbpedia.org/resource/>
DESCRIBE :Benin
-- or just --
DESCRIBE <http://dbpedia.org/resource/Benin>
You can get the output in various other serializations, including N-triples.
Related
I guess I am stuck at the basics with SPARQL. Can someone help ?
I simply wnat to filter all subjects containing "Mountain" of an RDS database.
Prefix lgdr:<http://linkedgeodata.org/triplify/> Prefix lgdo:<http://linkedgeodata.org/ontology/>
Select * where {
?s ?p ?o .
filter (contains(?s, "Mountain"))
} Limit 1000
The query leads to an error:
Virtuoso 22023 Error SL001: The SPARQL 1.1 function CONTAINS() needs a string value as first argument
You can get it to "work" using:
Prefix lgdr:<http://linkedgeodata.org/triplify/> Prefix lgdo:<http://linkedgeodata.org/ontology/>
Select * where {
?s ?p ?o .
filter (contains(str(?s), "Mountain"))
} Limit 1000
Note the additional str in the query.
However, that results in
Virtuoso S1T00 Error SR171: Transaction timed out
and I am not sure how to deal with that.
But in principle in works: When you use
Limit 1
you get
s p o
http://linkedgeodata.org/ontology/MountainRescue http://www.w3.org/1999/02/22-rdf-syntax-ns#type http://www.w3.org/2002/07/owl#Class
I am a newbie to SPARQL. Though I have read some materials about RDF and SPARQL, I still cannot understand the meaning of the mysterious SPARQL query on the LinkedGeoData SPARQL endpoint
Prefix lgdr:<http://linkedgeodata.org/triplify/>
Prefix lgdo:<http://linkedgeodata.org/ontology/>
Select * { ?s ?p ?o . }
Limit 1000
What does this oversimplified where condition ?s ?p ?o mean?
The query you ask about will return 1,000 triples from the endpoint with no filter or condition applied i.e. {?s ?p ?o. } will match any triple.
It's similar to SELECT * FROM aView in SQL if aView was a union of all, or most, of the tables in a SQL database.
In the following query I'm trying to get a list of all entries ?s that include more than 3 objects for the predicate sctap:mentionedBy. However, I keep getting a malformed query error for this search. Does anyone see anything wrong with my query?
Thanks
SELECT ?s
WHERE {
?s sctap:mentionedBy ?o
FILTER (count(?o) > 3)
}
The sparql error says: "Aggregate expression not legal". I'm not sure what that means.
Does anyone see anything wrong with my query?
Sure. Just like the error message says, you're using an aggregate expression (count(?o)) where one isn't legal. You can see in the table of contents of SPARQL 1.1 Query Language what things are filter functions that you can use in a filter, and what things are aggregates, and where you can use each. You can also try parsing queries at sparql.org's query validator. For your query, it will give you the line and column numbers where something went wrong. It's at count(?o).
In this case, you're trying to count the number of ?o values for each s, which means that you need to group by ?s, and that your filter will need to be father out. E.g.,
select ?s where {
?s sctap:mentionedBy ?o
}
group by ?s
having (count(?o) > 3)
It's unlikely to make a difference in this case, but you probably only want to count distinct values of ?o, so you could also consider:
select ?s where {
?s sctap:mentionedBy ?o
}
group by ?s
having (count(distinct ?o) > 3)
i have to use ROWLKit
http://www.dis.uniroma1.it/quonto/?q=node/30
(1) can anybody suggest two sparql queries for the Pizza.owl ?
(2) is this query valid ?
PREFIX rdf: <http://www.w3.org/1999/02/22-rdf-syntax-ns#>
PREFIX pizza: <http://www.co-ode.org/ontologies/pizza/pizza.owl#>
SELECT *
WHERE { ?p rdf:type pizza:Pizza;
pizza:hasTopping ?t.
?t rdf:type pizza:TomatoTopping }
(3) if it is a valid query then: is the response an empty result?
SELECT DISTINCT *
WHERE {
?NombrePizza ?Relacion pizza:MushroomTopping .
?Relacion owl:inverseOf pizza:isToppingOf .
OPTIONAL {
?NombrePizza2 ?Relacion2 pizza:HamTopping .
?Relacion2 owl:inverseOf pizza:isToppingOf .
}
FILTER(?NombrePizza2 = ?NombrePizza)
}
(1) can anybody suggest two sparql queries for the Pizza.owl ?
Here are two examples:
SELECT * WHERE { ?s ?p ?o }
and:
SELECT ?class WHERE { ?class a owl:Class }
(2) is this query valid ?
Yes.
(3) if it is a valid query then: is the response an empty result?
I assume that you mean "if I query the RDF document that serialises the pizza ontology, is the response an empty result?". The answer is yes.
(2) appears to be a valid query
I don't understand part (3) of your question. (2) cannot be compared to a boolean since it returns a Result Set, if you want a boolean result then you need to use an ASK query. If an ASK query returns true then it means that there are solutions to the query in the data you are querying so it would not be an empty result.
I want to select a triple using SPARQL. To do it, i'm using following query:
SELECT count (*)
WHERE {?s ?p ?o}
FILTER (?s=http://kjkhlsa.net && ?p=http://lkasdjlkjas.com && ?o=Test)
As answer i get fully wrong triple :( subject ist not equal to "http://kjkhlsa.net", predicate is not equal to "http://lkasdjlkjas.com" and object ist also not equal to "Test". Can someone explain me, what I'm doing wrong :(
edit1:
I have put the query into php file:
$inst_query = 'SELECT * { <http://kjkhlsa.net> <http://lkasdjlkjas.com> "Test"}';
echo $inst_query;
The answer from the echo was "SELECT * { "Test"}". Then i tried it with WHERE:
$inst_query = 'SELECT * WHERE { <http://kjkhlsa.net> <http://lkasdjlkjas.com> "Test"}';
echo $inst_query;
Here was the answer "SELECT * WHERE { "Test"}"...so, i'm missing the URIs, but this seems for me as php issue and not sparql problem.
edit2:
I've put the query into SPARQL Query editor and i get the response "no result"....but I'm sure, that i have this triple.
In its current form the question is not very clear (see my comment above).
Since you are essentially trying to get triples matching a pattern, it is more efficient to use a graph pattern instead of FILTER. Many SPARQL implementations first match candidate triples by graph patterns and only then apply the FILTER expression. In essence, with a ?s ?p ?o graph pattern, you're doing a linear scan over all your triples.
So, here's something that should work, using graph patterns instead of FILTER.
SELECT * { <http://kjkhlsa.net> <http://lkasdjlkjas.com> "Test" }
Notes: I didn't include COUNT(*) which is not standard SPARQL. <> around URIs. "" around literal.
Try to use this :
SELECT count (*) as ?count
WHERE {
?s ?p ?o
FILTER (?s=<http://kjkhlsa.net> && ?p=<http://lkasdjlkjas.com> && ?o=Test)
}
The following query uses the count function to count the number of distinct URI(s) returned to the ?s variable.
SELECT ?s (COUNT (DISTINCT ?s) as ?count)
WHERE {?s ?p ?o}
FILTER (?o="Test")