I have a list of dbpedia URI's and I want to get some informations (categories, label) about each of them in one query:
SELECT ?category ?label where {
{
dbpedia:Financial_Times dcterms:subject ?category .
dbpedia:Financial_Times rdfs:label ?label .
FILTER ( lang(?label) = 'en' )
}
UNION
{
dbpedia:London dcterms:subject ?category .
dbpedia:London rdfs:label ?label .
FILTER ( lang(?label) = 'en' )
}
}
This query works fine, but I'd need to add the URI's themself into the result to be able identify which result row is for which URI.
you can do something like
SELECT distinct ?who ?category ?label where {
{
?who dcterms:subject ?category .
?who rdfs:label ?label .
FILTER ( lang(?label) = 'en' ).
FILTER(?who = dbpedia:Financial_Times or ?who = dbpedia:London )
}}
or use a trick like this
SELECT ?who ?category ?label where {
{
dbpedia:Financial_Times dcterms:subject ?category .
dbpedia:Financial_Times rdfs:label ?label .
FILTER ( lang(?label) = 'en' ).
VALUES ?who { dbpedia:Financial_Times}
}
UNION
{
dbpedia:London dcterms:subject ?category .
dbpedia:London rdfs:label ?label .
FILTER ( lang(?label) = 'en' ) .
VALUES ?who { dbpedia:London }
}}
the second one probably is faster but needs SPARQL 1.1
Related
Could anyone please help me with this code. It doesn't come out with any output but with no error.
q = """SELECT DISTINCT ?label ?abstract ?director ?starring
WHERE {
<http://dbpedia.org/resource/Seven_Beauties> rdfs:label ?label.
?label dbo:abstract ?abstract.
?label dbo:director ?director.
?label dbo:starring ?starring.
FILTER (lang(?label) = "en")
FILTER (lang(?abstract) = "en")
}"""
from textwrap import wrap
for result in query(q):
print(result['label'],
"\n----Director----\n",result['director'],
"\n----Starring----\n",result['starring'],
"\n----Abstract----\n",
"\n".join(wrap(result['abstract'])))
<http://dbpedia.org/resource/Seven_Beauties> is the entity which has a Label, Abstract, Director, Star, etc.
The Label of that entity is a literal, and it has no Director, Star, Abstract, etc.
Try changing your query to --
SELECT DISTINCT ?label ?abstract ?director ?starring
WHERE
{
<http://dbpedia.org/resource/Seven_Beauties>
rdfs:label ?label ;
dbo:abstract ?abstract ;
dbo:director ?director ;
dbo:starring ?starring
FILTER ( lang(?label) = "en" )
FILTER ( lang(?abstract) = "en" )
}
-- and see results from DBpedia.
DBpedia Live delivers nothing for the query above -- because the descriptions of films vary. Making some things optional will get you more results on DBpedia Live --
SELECT DISTINCT ?label ?abstract ?director ?starring
WHERE
{
<http://dbpedia.org/resource/Seven_Beauties> rdfs:label ?label
OPTIONAL { <http://dbpedia.org/resource/Seven_Beauties> dbo:abstract ?abstract }
OPTIONAL { <http://dbpedia.org/resource/Seven_Beauties> dbo:director ?director }
OPTIONAL { <http://dbpedia.org/resource/Seven_Beauties> dbo:starring ?starring }
FILTER ( lang(?label) = "en" )
FILTER ( lang(?abstract) = "en" )
}
Trying to execute some queries but when searching for foaf:name resultset is empty. Here's my code:
SELECT DISTINCT ?uri ?string
WHERE {
?uri rdf:type ?x.
?uri foaf:name 'Cavallo domestico'#it .
OPTIONAL { ?uri rdfs:label ?string . FILTER (lang(?string) = 'it') }
}
page exist http://it.dbpedia.org/resource/Equus_caballus/html
Apparently seems it's not related with languages different than english but with foaf:name request. If I execute following, retrieving generic foaf:givenName, it works:
SELECT DISTINCT ?uri ?string
WHERE {
?uri rdf:type ?x.
?uri foaf:givenName 'Jimmy'#en .
OPTIONAL { ?uri rdfs:label ?string . FILTER (lang(?string) = 'en') }
}
I think this wasn't working when I first mentioned that it didn't in a comment, but, as AKSW points out, this seems to be working now. The rdfs:label property has the article titles in various languages, not the foaf:name, so you can do this to get the types of Horse:
select ?x ?type {
?x a ?type ;
rdfs:label "Equus caballus"#it
}
SPARQL results
I have two SPARQL updates.First one:
INSERT
{ GRAPH <[http://example/bookStore2]> { ?book ?p ?v } }
WHERE
{ GRAPH <[http://example/bookStore]>
{ ?book dc:date ?date .
FILTER ( ?date > "1970-01-01T00:00:00-02:00"^^xsd:dateTime )
?book ?p ?v
} }
Second:
INSERT
{ GRAPH <[http://example/bookStore2]> { ?book ?p ?v } }
WHERE
{ GRAPH <[http://example/bookStore3]>
{ ?book dc:date ?date .
FILTER ( ?date > "1980-01-01T00:00:00-02:00"^^xsd:dateTime )
?book ?p ?v
} }
Can i combine them with the UNION operator? And if yes, is it an equivalent result? Is it possible to use UNION in SPARQL updates such as in "Select"?
AndyS's answer is correct; you can combine them, and the description of UNION is found in section 7 Matching Alternatives of the SPARQL specification. The combined query would be:
INSERT {
GRAPH <[http://example/bookStore2]> { ?book ?p ?v }
}
WHERE{
{
GRAPH <[http://example/bookStore]> {
?book dc:date ?date .
FILTER ( ?date > "1970-01-01T00:00:00-02:00"^^xsd:dateTime )
?book ?p ?v
}
}
UNION
{
GRAPH <[http://example/bookStore3]> {
?book dc:date ?date .
FILTER ( ?date > "1980-01-01T00:00:00-02:00"^^xsd:dateTime )
?book ?p ?v
}
}
}
In this particular case where the patterns are so similar, you could also just abstract out the differing parts with VALUES:
INSERT {
GRAPH <[http://example/bookStore2]> { ?book ?p ?v }
}
WHERE{
values (?graph ?startDate) {
(<[http://example/bookStore]> "1970-01-01T00:00:00-02:00"^^xsd:dateTime)
(<[http://example/bookStore3]> "1980-01-01T00:00:00-02:00"^^xsd:dateTime)
}
GRAPH ?graph {
?book dc:date ?date .
FILTER ( ?date > ?startDate )
?book ?p ?v
}
}
The WHERE clause is the same as SPARQL Query - you can use UNION.
I'd like to do something like
{
SELECT ?page, "A" AS ?type WHERE
{
?s rdfs:label "Microsoft"#en;
foaf:page ?page
}
}
UNION
{
SELECT ?page, "B" AS ?type WHERE
{
?s rdfs:label "Apple"#en;
foaf:page ?page
}
}
But this gives a syntax error. How can I union two select queries in SPARQL?
You can union them like this:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX foaf: <http://xmlns.com/foaf/0.1/>
SELECT * WHERE
{
{
SELECT ?page ("A" AS ?type) WHERE
{
?s rdfs:label "Microsoft"#en;
foaf:page ?page
}
}
UNION
{
SELECT ?page ("B" AS ?type) WHERE
{
?s rdfs:label "Apple"#en;
foaf:page ?page
}
}
}
(check with the SPARQL validator)
However I don't think you need sub queries at all for this case. For example:
PREFIX rdfs: <http://www.w3.org/2000/01/rdf-schema#>
PREFIX foaf: <http://xmlns.com/foaf/0.1/>
SELECT ?page ?type WHERE
{
?s foaf:page ?page .
{ ?s rdfs:label "Microsoft"#en . BIND ("A" as ?type) }
UNION
{ ?s rdfs:label "Apple"#en . BIND ("B" as ?type) }
}
Based on #user205512's answer, here's one that works on Virtuoso:
SELECT * {
?s foaf:page ?page .
{
SELECT ?page ("A" AS ?type) {
?s rdfs:label "Microsoft"#en;
foaf:page ?page
}
} UNION {
SELECT ?page ("B" AS ?type) {
?s rdfs:label "Apple"#en;
foaf:page ?page
}
}
}
The trick was just do add an additional ?s foaf:page ?page triple outside of the UNION. This is obviously redundant, but it seems to avoid the Virtuoso bug, which is apparently caused when you have a “naked” UNION with subqueries.
I want to get triples about cities, which are from certain country. How can I do that?
I tried:
CONSTRUCT { ?c rdfs:label ?name . ?c rdfs:comment ?desc }
WHERE {
?c dbpprop:wikiPageUsesTemplate <http://dbpedia.org/resource/Template:Infobox_settlement> .
?c rdfs:label ?name .
?c rdfs:comment ?desc .
?c <http://dbpedia.org/ontology/country> ?country . ?country a <http://dbpedia.org/resource/CountryName>
FILTER ( lang(?name) = "en" && lang(?desc) = "en" )
}
but no luck :/ how can i do this?
CONSTRUCT { ?c rdfs:label ?name }
WHERE {
?c dbpprop:wikiPageUsesTemplate <http://dbpedia.org/resource/Template:Infobox_settlement> .
?c rdfs:label ?name .
?c dbpedia-owl:country <http://dbpedia.org/resource/Country> .
OPTIONAL { ?c dbpedia-owl:areaCode ?areacode }
FILTER ( lang(?name) = "pl" && ?population > 5000)
}
Hope it will help :)